{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0001","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c4a9085190b32e2c35e968e6873eff1ea405a20183d2e2146ea76363beff9b3","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275001,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-12,-3,-5],[-2,12,-6]],\"vector\":[-4,-2,-3]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-12, -3, -5], [-2, 12, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-4, -2, -3]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-12, -3, -5], [-2, 12, -6]], [-4, -2, -3], [69, 2])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-12, -3, -5], [-2, 12, -6]], [-4, -2, -3], [69, 2])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0002","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"648c3ad0ffd72d1f8b74c8becc399426015d55292746de3b0af84ca4be7edfbb","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275002,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-11,-2,-2],[-1,11,-1]],\"vector\":[-3,-1,-2]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-11, -2, -2], [-1, 11, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-3, -1, -2]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-11, -2, -2], [-1, 11, -1]], [-3, -1, -2], [39, -6])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-11, -2, -2], [-1, 11, -1]], [-3, -1, -2], [39, -6])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0003","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43cea9b2354c35136bdade991eb2cfb8a95dfa8db560ac8766ed9ff5ba25b36f","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275003,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-10,-1,1],[0,10,4]],\"vector\":[-2,0,-1]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-10, -1, 1], [0, 10, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-2, 0, -1]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-10, -1, 1], [0, 10, 4]], [-2, 0, -1], [19, -4])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-10, -1, 1], [0, 10, 4]], [-2, 0, -1], [19, -4])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0004","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78e54713612c6dcf898280d05d19221974a7f73ecf6e944f1095879819ce9651","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275004,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-9,0,4],[1,9,-4]],\"vector\":[-1,1,0]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-9, 0, 4], [1, 9, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-1, 1, 0]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-9, 0, 4], [1, 9, -4]], [-1, 1, 0], [9, 8])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-9, 0, 4], [1, 9, -4]], [-1, 1, 0], [9, 8])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0005","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f026a037e7269a9cf896c4f6dc08aa8090ce5e7e6de0647e45474fa34c201b0d","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275005,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-8,1,-4],[2,8,1]],\"vector\":[0,2,1]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-8, 1, -4], [2, 8, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[0, 2, 1]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-8, 1, -4], [2, 8, 1]], [0, 2, 1], [-2, 17])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-8, 1, -4], [2, 8, 1]], [0, 2, 1], [-2, 17])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0006","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42ca8458e8eb099885890c172b2bd50aada3ad708d6997908f80102659921cca","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275006,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-7,2,-1],[-2,7,6]],\"vector\":[1,-2,2]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-7, 2, -1], [-2, 7, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[1, -2, 2]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-7, 2, -1], [-2, 7, 6]], [1, -2, 2], [-13, -4])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-7, 2, -1], [-2, 7, 6]], [1, -2, 2], [-13, -4])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0007","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f53c8a982610385a64cf078923ae283168d2180dff0194e89fedfef65e22a910","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275007,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-6,3,2],[-1,6,-2]],\"vector\":[2,-1,3]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-6, 3, 2], [-1, 6, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[2, -1, 3]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-6, 3, 2], [-1, 6, -2]], [2, -1, 3], [-9, -14])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-6, 3, 2], [-1, 6, -2]], [2, -1, 3], [-9, -14])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0008","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"672497a1d23752483bf338e347bb465e8d6a26bdb505e61afdf31030b16c6e0f","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275008,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-5,-3,5],[0,5,3]],\"vector\":[3,0,-3]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-5, -3, 5], [0, 5, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[3, 0, -3]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-5, -3, 5], [0, 5, 3]], [3, 0, -3], [-30, -9])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-5, -3, 5], [0, 5, 3]], [3, 0, -3], [-30, -9])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0009","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b0e786da643ed95aa4a93bac918bdf62676848de7e0272336c64e94fa02e965","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275009,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-4,-2,-3],[1,4,-5]],\"vector\":[4,1,-2]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-4, -2, -3], [1, 4, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[4, 1, -2]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-4, -2, -3], [1, 4, -5]], [4, 1, -2], [-12, 18])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-4, -2, -3], [1, 4, -5]], [4, 1, -2], [-12, 18])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0010","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"337394caa2628e675ef9d9de996fc4e34043ec029881a5e826082fe67fcf6222","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275010,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-3,-1,0],[2,3,0]],\"vector\":[-4,2,-1]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-3, -1, 0], [2, 3, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-4, 2, -1]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-3, -1, 0], [2, 3, 0]], [-4, 2, -1], [10, -2])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-3, -1, 0], [2, 3, 0]], [-4, 2, -1], [10, -2])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0011","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7647ceb3c92a2c14ea4cd1af0f249b32fd3b497b031800cb0ed5586a7cdbcaaa","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275011,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-2,0,3],[-2,2,5]],\"vector\":[-3,-2,0]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-2, 0, 3], [-2, 2, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-3, -2, 0]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-2, 0, 3], [-2, 2, 5]], [-3, -2, 0], [6, 2])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-2, 0, 3], [-2, 2, 5]], [-3, -2, 0], [6, 2])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0012","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8533b50a679bff2c4e96def92df7e52bbb4de641051c29de034e9dad803801ba","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275012,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[-1,1,-5],[-1,1,-3]],\"vector\":[-2,-1,1]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[-1, 1, -5], [-1, 1, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-2, -1, 1]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[-1, 1, -5], [-1, 1, -3]], [-2, -1, 1], [-4, -2])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-1, 1, -5], [-1, 1, -3]], [-2, -1, 1], [-4, -2])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0013","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3eb95463e994d6e5faee476d336edae30738aa29492f57cf4da3a6c4052e7dba","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275013,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[0,2,-2],[0,0,2]],\"vector\":[-1,0,2]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[0, 2, -2], [0, 0, 2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-1, 0, 2]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[0, 2, -2], [0, 0, 2]], [-1, 0, 2], [-4, 4])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[0, 2, -2], [0, 0, 2]], [-1, 0, 2], [-4, 4])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0014","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e0bed7e9088447098d18af21ae9558a672af72900821e07d80b4b66a8a0cae7","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275014,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[1,3,1],[1,-1,-6]],\"vector\":[0,1,3]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[1, 3, 1], [1, -1, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[0, 1, 3]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[1, 3, 1], [1, -1, -6]], [0, 1, 3], [6, -19])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[1, 3, 1], [1, -1, -6]], [0, 1, 3], [6, -19])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0015","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"edbe65a0522ed38f4122672a6a185ad6717a3a539a65ed3b53792c7a8150fe4e","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275015,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[2,-3,4],[2,-2,-1]],\"vector\":[1,2,-3]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[2, -3, 4], [2, -2, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[1, 2, -3]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[2, -3, 4], [2, -2, -1]], [1, 2, -3], [-16, 1])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[2, -3, 4], [2, -2, -1]], [1, 2, -3], [-16, 1])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0016","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f685618a1b2b433332933d192722a0f504a7880c86a9c334841c1b064843d21","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275016,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[3,-2,-4],[-2,-3,4]],\"vector\":[2,-2,-2]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[3, -2, -4], [-2, -3, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[2, -2, -2]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[3, -2, -4], [-2, -3, 4]], [2, -2, -2], [18, -6])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[3, -2, -4], [-2, -3, 4]], [2, -2, -2], [18, -6])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0017","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a403c8a643bb1a03d2687760c57adac9c2fed5ff3293ffa60f508efe809d7f1b","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275017,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[4,-1,-1],[-1,-4,-4]],\"vector\":[3,-1,-1]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[4, -1, -1], [-1, -4, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[3, -1, -1]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[4, -1, -1], [-1, -4, -4]], [3, -1, -1], [14, 5])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[4, -1, -1], [-1, -4, -4]], [3, -1, -1], [14, 5])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0018","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"211723cf34c9165b64c17fec063db1c19778bd58fdda96d8c30e8fe05a9bfd73","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275018,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[5,0,2],[0,-5,1]],\"vector\":[4,0,0]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[5, 0, 2], [0, -5, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[4, 0, 0]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[5, 0, 2], [0, -5, 1]], [4, 0, 0], [20, 0])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[5, 0, 2], [0, -5, 1]], [4, 0, 0], [20, 0])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0019","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"290f0e132347731def5745540b868f9210c52b0531c85eccb3753146902b061d","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275019,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[6,1,5],[1,-6,6]],\"vector\":[-4,1,1]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[6, 1, 5], [1, -6, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-4, 1, 1]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[6, 1, 5], [1, -6, 6]], [-4, 1, 1], [-18, -4])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[6, 1, 5], [1, -6, 6]], [-4, 1, 1], [-18, -4])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0020","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2a0e502b1b186448aa7866af65c0897335e7166316718631c3c42905b8b214c","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275020,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[7,2,-3],[2,-7,-2]],\"vector\":[-3,2,2]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[7, 2, -3], [2, -7, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-3, 2, 2]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[7, 2, -3], [2, -7, -2]], [-3, 2, 2], [-23, -24])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[7, 2, -3], [2, -7, -2]], [-3, 2, 2], [-23, -24])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0021","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cec96fefbb13a06ddbd9318930716d3abfb0ab4f02c4daa72bc6b96472f195c4","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275021,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[8,3,0],[-2,-8,3]],\"vector\":[-2,-2,3]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[8, 3, 0], [-2, -8, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-2, -2, 3]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[8, 3, 0], [-2, -8, 3]], [-2, -2, 3], [-22, 29])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[8, 3, 0], [-2, -8, 3]], [-2, -2, 3], [-22, 29])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0022","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33dbba9107ccc003046e7a4a0d52905c76c98a481d0f56e3ea29fbe3f1ce525a","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275022,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[9,-3,3],[-1,-9,-5]],\"vector\":[-1,-1,-3]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[9, -3, 3], [-1, -9, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[-1, -1, -3]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[9, -3, 3], [-1, -9, -5]], [-1, -1, -3], [-15, 25])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[9, -3, 3], [-1, -9, -5]], [-1, -1, -3], [-15, 25])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0023","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe8cdd44252907692f80a7c4c8db1231bccb24aa36cd96ed8a56996573fed28d","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275023,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[10,-2,-5],[0,-10,0]],\"vector\":[0,0,-2]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[10, -2, -5], [0, -10, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[0, 0, -2]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[10, -2, -5], [0, -10, 0]], [0, 0, -2], [10, 0])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[10, -2, -5], [0, -10, 0]], [0, 0, -2], [10, 0])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0024","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e18808596ab7c8e4d63da99d90ea4b4abd9dc43119379cc862f2371a4f6eaff8","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275024,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[11,-1,-2],[1,-11,5]],\"vector\":[1,1,-1]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[11, -1, -2], [1, -11, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[1, 1, -1]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[11, -1, -2], [1, -11, 5]], [1, 1, -1], [12, -15])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[11, -1, -2], [1, -11, 5]], [1, 1, -1], [12, -15])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0025","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f429f717eb8ab13791068a47bc328cdcd7ff55ffcbe5a377486c127fd0137e1","family_id":"qf-mod27-linearmap-matrix-evaluate","seed":202610275025,"semantic_template":"qf-mod27-linearmap-matrix-evaluate/{\"matrix\":[[12,0,1],[2,-12,-3]],\"vector\":[2,2,0]}"},"fields":{"title":"Evaluate a Possibly Rectangular Real Linear Map","stem":"Use the complete rectangular matrix A=[[12, 0, 1], [2, -12, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. vector=[2, 2, 0]. Give (matrix,vector,image).","answer":{"kind":"expression","value":"([[12, 0, 1], [2, -12, -3]], [2, 2, 0], [24, -20])","explanation":"Take each row dot product with the input vector, retaining the row/column direction and original order. The input and output dimensions need not be equal. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[12, 0, 1], [2, -12, -3]], [2, 2, 0], [24, -20])."},"skills":["Evaluate a Possibly Rectangular Real Linear Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0026","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72c6fa826ad61bfa1e540bb360c4c52a1c669968244cad3ec6fb4a738103da38","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275026,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[0,0],\"matrix\":[[-12,-3,-5],[-2,12,-6]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-12, -3, -5], [-2, 12, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[0, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-12, -3, -5], [-2, 12, -6]], [0, 0], [0, 0], Yes, matrix_linear_certificate)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-12, -3, -5], [-2, 12, -6]], [0, 0], [0, 0], Yes, matrix_linear_certificate)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0027","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"735252e5bcb8cca8319c8cf4371a516dd8384d5c104cec23a25e8b53c152ca75","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275027,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[2,-2],\"matrix\":[[-11,-2,-2],[-1,11,-1]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-11, -2, -2], [-1, 11, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[2, -2]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-11, -2, -2], [-1, 11, -1]], [2, -2], [2, -2], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-11, -2, -2], [-1, 11, -1]], [2, -2], [2, -2], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0028","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9e2e69d4fa8964f4b8558c17908ba599f6a9d51729fc82c92f68af6a9b33526","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275028,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[3,-1],\"matrix\":[[-10,-1,1],[0,10,4]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-10, -1, 1], [0, 10, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[3, -1]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-10, -1, 1], [0, 10, 4]], [3, -1], [3, -1], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-10, -1, 1], [0, 10, 4]], [3, -1], [3, -1], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0029","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0807c4e57dfa1cdcb0c200e5ce59d250398bd67f730dd75edbfddcfd1c001d8c","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275029,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[4,0],\"matrix\":[[-9,0,4],[1,9,-4]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-9, 0, 4], [1, 9, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[4, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-9, 0, 4], [1, 9, -4]], [4, 0], [4, 0], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-9, 0, 4], [1, 9, -4]], [4, 0], [4, 0], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0030","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"309203f5c9a2492d6aa12c1d5cf70ac85b8934cca2659197b7d5defeb0bd3444","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275030,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[0,0],\"matrix\":[[-8,1,-4],[2,8,1]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-8, 1, -4], [2, 8, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[0, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-8, 1, -4], [2, 8, 1]], [0, 0], [0, 0], Yes, matrix_linear_certificate)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-8, 1, -4], [2, 8, 1]], [0, 0], [0, 0], Yes, matrix_linear_certificate)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0031","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a30f249d841badfa2fab9dbb07f5133d2b004850d3f1298832aca14b24efdfb","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275031,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[1,2],\"matrix\":[[-7,2,-1],[-2,7,6]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-7, 2, -1], [-2, 7, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[1, 2]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-7, 2, -1], [-2, 7, 6]], [1, 2], [1, 2], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-7, 2, -1], [-2, 7, 6]], [1, 2], [1, 2], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0032","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6ff0bb14c9d0719e7b661d8e09b3ddb707eeafe0d80dd5266a19bc4e60464d2","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275032,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[2,3],\"matrix\":[[-6,3,2],[-1,6,-2]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-6, 3, 2], [-1, 6, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[2, 3]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-6, 3, 2], [-1, 6, -2]], [2, 3], [2, 3], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-6, 3, 2], [-1, 6, -2]], [2, 3], [2, 3], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0033","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"020c2d859472cebbbd8304703fc93a2b4eb5ca8cfab96821503453121cded67b","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275033,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[3,-3],\"matrix\":[[-5,-3,5],[0,5,3]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-5, -3, 5], [0, 5, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[3, -3]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-5, -3, 5], [0, 5, 3]], [3, -3], [3, -3], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-5, -3, 5], [0, 5, 3]], [3, -3], [3, -3], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0034","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c82bed52295242b986cb4f31ea2b1626658ce1a1c61760192ce9bac77e024d6","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275034,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[0,0],\"matrix\":[[-4,-2,-3],[1,4,-5]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-4, -2, -3], [1, 4, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[0, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-4, -2, -3], [1, 4, -5]], [0, 0], [0, 0], Yes, matrix_linear_certificate)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-4, -2, -3], [1, 4, -5]], [0, 0], [0, 0], Yes, matrix_linear_certificate)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0035","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ddc617a7993a784f8f7bd1cafed1ccb44fd3a4c04c6b10f8ecde5042f37264ce","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275035,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[5,-1],\"matrix\":[[-3,-1,0],[2,3,0]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-3, -1, 0], [2, 3, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[5, -1]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-3, -1, 0], [2, 3, 0]], [5, -1], [5, -1], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-3, -1, 0], [2, 3, 0]], [5, -1], [5, -1], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0036","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdf2bb5b00b91e9402dec974399d632b5536b371c244a4f9e06f8f7be24a5566","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275036,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[1,0],\"matrix\":[[-2,0,3],[-2,2,5]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-2, 0, 3], [-2, 2, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[1, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-2, 0, 3], [-2, 2, 5]], [1, 0], [1, 0], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-2, 0, 3], [-2, 2, 5]], [1, 0], [1, 0], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0037","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02e640e99416243f6f95abe41299a9ac864caba61d5c8c76c0889589fd4f1289","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275037,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[2,1],\"matrix\":[[-1,1,-5],[-1,1,-3]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[-1, 1, -5], [-1, 1, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[2, 1]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[-1, 1, -5], [-1, 1, -3]], [2, 1], [2, 1], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[-1, 1, -5], [-1, 1, -3]], [2, 1], [2, 1], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0038","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"567276bf4d69f375c6f5c93678284407ea76d531a9a918309e6f0b44a39e3319","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275038,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[0,0],\"matrix\":[[0,2,-2],[0,0,2]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[0, 2, -2], [0, 0, 2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[0, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[0, 2, -2], [0, 0, 2]], [0, 0], [0, 0], Yes, matrix_linear_certificate)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[0, 2, -2], [0, 0, 2]], [0, 0], [0, 0], Yes, matrix_linear_certificate)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0039","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42404b2f0e89a8ce54162daf513f47b4c4edb4f50e9984c49bdb31ea533c29a5","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275039,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[4,3],\"matrix\":[[1,3,1],[1,-1,-6]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[1, 3, 1], [1, -1, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[4, 3]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[1, 3, 1], [1, -1, -6]], [4, 3], [4, 3], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[1, 3, 1], [1, -1, -6]], [4, 3], [4, 3], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0040","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e951498bd5ba93b6cc96282c62d0585605a4e3e42407b402817882e407aed178","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275040,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[5,-3],\"matrix\":[[2,-3,4],[2,-2,-1]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[2, -3, 4], [2, -2, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[5, -3]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[2, -3, 4], [2, -2, -1]], [5, -3], [5, -3], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[2, -3, 4], [2, -2, -1]], [5, -3], [5, -3], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0041","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4da0228d3e79998d7220e8098584d651c9539e75624e0433d20b0aa84bf41a7d","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275041,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[1,-2],\"matrix\":[[3,-2,-4],[-2,-3,4]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[3, -2, -4], [-2, -3, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[1, -2]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[3, -2, -4], [-2, -3, 4]], [1, -2], [1, -2], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[3, -2, -4], [-2, -3, 4]], [1, -2], [1, -2], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0042","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ccc01ddf9374e3667f2114b3698732e6cb2ba599c3f201931f176c7cf559f9b","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275042,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[0,0],\"matrix\":[[4,-1,-1],[-1,-4,-4]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[4, -1, -1], [-1, -4, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[0, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[4, -1, -1], [-1, -4, -4]], [0, 0], [0, 0], Yes, matrix_linear_certificate)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[4, -1, -1], [-1, -4, -4]], [0, 0], [0, 0], Yes, matrix_linear_certificate)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0043","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1718520e12e3fc71da114b49fef0c4905d6de1155ccb8b8e8bb81c626444ee73","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275043,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[3,0],\"matrix\":[[5,0,2],[0,-5,1]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[5, 0, 2], [0, -5, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[3, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[5, 0, 2], [0, -5, 1]], [3, 0], [3, 0], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[5, 0, 2], [0, -5, 1]], [3, 0], [3, 0], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0044","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7271311b4596145acc2009f1a2d2ba0ba6c02b588884aae3883010e0829e1f84","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275044,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[4,1],\"matrix\":[[6,1,5],[1,-6,6]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[6, 1, 5], [1, -6, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[4, 1]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[6, 1, 5], [1, -6, 6]], [4, 1], [4, 1], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[6, 1, 5], [1, -6, 6]], [4, 1], [4, 1], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0045","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ab170b8474761092d3a5e03ab495c05c8a0090c7537aff50ed2d9bf76c2bc23c","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275045,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[5,2],\"matrix\":[[7,2,-3],[2,-7,-2]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[7, 2, -3], [2, -7, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[5, 2]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[7, 2, -3], [2, -7, -2]], [5, 2], [5, 2], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[7, 2, -3], [2, -7, -2]], [5, 2], [5, 2], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0046","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dbc01be69ffbe6a80a9cede62b17f494c9890f89bfcfdaa4aa546ec05b8e3c0","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275046,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[0,0],\"matrix\":[[8,3,0],[-2,-8,3]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[8, 3, 0], [-2, -8, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[0, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[8, 3, 0], [-2, -8, 3]], [0, 0], [0, 0], Yes, matrix_linear_certificate)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[8, 3, 0], [-2, -8, 3]], [0, 0], [0, 0], Yes, matrix_linear_certificate)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0047","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e78103ab0c212437c3be2e20d6a9aea505182d2812ef22b9d628ac7e6570ef3f","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275047,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[2,-3],\"matrix\":[[9,-3,3],[-1,-9,-5]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[9, -3, 3], [-1, -9, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[2, -3]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[9, -3, 3], [-1, -9, -5]], [2, -3], [2, -3], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[9, -3, 3], [-1, -9, -5]], [2, -3], [2, -3], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0048","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33d61a32be95881bc3a8bd19a3da084bf721b5f131a7cf62d2fc578cc8f6da76","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275048,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[3,-2],\"matrix\":[[10,-2,-5],[0,-10,0]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[10, -2, -5], [0, -10, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[3, -2]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[10, -2, -5], [0, -10, 0]], [3, -2], [3, -2], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[10, -2, -5], [0, -10, 0]], [3, -2], [3, -2], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0049","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e949a63a054365aa4bc672c3b13563b67909027a9bab767a780615790fdd12d","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275049,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[4,-1],\"matrix\":[[11,-1,-2],[1,-11,5]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[11, -1, -2], [1, -11, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[4, -1]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[11, -1, -2], [1, -11, 5]], [4, -1], [4, -1], No, nonzero_translation_is_not_linear)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[11, -1, -2], [1, -11, 5]], [4, -1], [4, -1], No, nonzero_translation_is_not_linear)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0050","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1505244e4ee16a88c7364f6f2005ab02ec5ad365507cb7cd6918399fa8118a4d","family_id":"qf-mod27-linearmap-affine-qualification","seed":202610275050,"semantic_template":"qf-mod27-linearmap-affine-qualification/{\"bias\":[0,0],\"matrix\":[[12,0,1],[2,-12,-3]]}"},"fields":{"title":"A Nonzero Affine Translation Fails the Zero-Image Obligation","stem":"Use the complete rectangular matrix A=[[12, 0, 1], [2, -12, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. The actual map is T(v)=A*v+bias, with bias=[0, 0]. Give (matrix,bias,T_zero,linear?,boundary). A zero bias yields matrix_linear_certificate; a nonzero bias yields nonzero_translation_is_not_linear.","answer":{"kind":"expression","value":"([[12, 0, 1], [2, -12, -3]], [0, 0], [0, 0], Yes, matrix_linear_certificate)","explanation":"Evaluate T at the zero vector in the input space. A nonzero bias immediately refutes linearity; a zero bias leaves the universal matrix formula. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([[12, 0, 1], [2, -12, -3]], [0, 0], [0, 0], Yes, matrix_linear_certificate)."},"skills":["A Nonzero Affine Translation Fails the Zero-Image Obligation"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0051","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72dee17eddeb31f524367c25506adae54a23afa7220622896ac853f1a37ab39b","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275051,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[0,0],\"matrix\":[[-12,-3,-5],[-2,12,-6]],\"u\":[1,-2,-3],\"v\":[-5,1,-2]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-12, -3, -5], [-2, 12, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[0, 0], u=[1, -2, -3], v=[-5, 1, -2]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -2, -3], [-5, 1, -2], [0, 0], [76, 26], [76, 26], [0, 0], Yes)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -2, -3], [-5, 1, -2], [0, 0], [76, 26], [76, 26], [0, 0], Yes)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0052","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6bfe654fd6290114890789ec9c0dbd279bc3634fe10f30640321f1396697998f","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275052,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[2,-2],\"matrix\":[[-11,-2,-2],[-1,11,-1]],\"u\":[1,-1,-2],\"v\":[-4,1,-1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-11, -2, -2], [-1, 11, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[2, -2], u=[1, -1, -2], v=[-4, 1, -1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -1, -2], [-4, 1, -1], [2, -2], [41, 4], [43, 2], [-2, 2], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -1, -2], [-4, 1, -1], [2, -2], [41, 4], [43, 2], [-2, 2], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0053","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5105ee7d52fffc4e0d3a5f38a84c992ccf39fd8ab35bc514ecf4c4ea124f7a74","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275053,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[3,-1],\"matrix\":[[-10,-1,1],[0,10,4]],\"u\":[1,0,-1],\"v\":[-3,1,0]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-10, -1, 1], [0, 10, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[3, -1], u=[1, 0, -1], v=[-3, 1, 0]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 0, -1], [-3, 1, 0], [3, -1], [21, 5], [24, 4], [-3, 1], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 0, -1], [-3, 1, 0], [3, -1], [21, 5], [24, 4], [-3, 1], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0054","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e043e789071b09e6639dd548a7baba0f85d7fdea189e3837a7e93c2b870de1a","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275054,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[4,0],\"matrix\":[[-9,0,4],[1,9,-4]],\"u\":[1,1,0],\"v\":[-2,1,1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-9, 0, 4], [1, 9, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[4, 0], u=[1, 1, 0], v=[-2, 1, 1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 1, 0], [-2, 1, 1], [4, 0], [17, 13], [21, 13], [-4, 0], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 1, 0], [-2, 1, 1], [4, 0], [17, 13], [21, 13], [-4, 0], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0055","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ce97ca49042c71fd6419e37ae61a9293e4f86fdf582f3ff3824675c1930d9b3","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275055,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[0,0],\"matrix\":[[-8,1,-4],[2,8,1]],\"u\":[1,2,1],\"v\":[-1,1,-2]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-8, 1, -4], [2, 8, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[0, 0], u=[1, 2, 1], v=[-1, 1, -2]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 2, 1], [-1, 1, -2], [0, 0], [7, 23], [7, 23], [0, 0], Yes)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 2, 1], [-1, 1, -2], [0, 0], [7, 23], [7, 23], [0, 0], Yes)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0056","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"491317310fa63ba3e6de82620b2b4aee29477bf0f81394953de08a1916eb6c2f","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275056,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[1,2],\"matrix\":[[-7,2,-1],[-2,7,6]],\"u\":[1,-2,2],\"v\":[0,1,-1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-7, 2, -1], [-2, 7, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[1, 2], u=[1, -2, 2], v=[0, 1, -1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -2, 2], [0, 1, -1], [1, 2], [-9, -1], [-8, 1], [-1, -2], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -2, 2], [0, 1, -1], [1, 2], [-9, -1], [-8, 1], [-1, -2], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0057","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61f5942c02ca140693b43a3a2f8bdc9fb5ea04493bbe79f0e0aa972a2824e98d","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275057,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[2,3],\"matrix\":[[-6,3,2],[-1,6,-2]],\"u\":[1,-1,3],\"v\":[1,1,0]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-6, 3, 2], [-1, 6, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[2, 3], u=[1, -1, 3], v=[1, 1, 0]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -1, 3], [1, 1, 0], [2, 3], [-4, -5], [-2, -2], [-2, -3], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -1, 3], [1, 1, 0], [2, 3], [-4, -5], [-2, -2], [-2, -3], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0058","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b4c90c34367b3e30184473be76074bd2eb00d3bed23784f53594caa14dca08b","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275058,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[3,-3],\"matrix\":[[-5,-3,5],[0,5,3]],\"u\":[1,0,-3],\"v\":[2,1,1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-5, -3, 5], [0, 5, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[3, -3], u=[1, 0, -3], v=[2, 1, 1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 0, -3], [2, 1, 1], [3, -3], [-25, -4], [-22, -7], [-3, 3], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 0, -3], [2, 1, 1], [3, -3], [-25, -4], [-22, -7], [-3, 3], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0059","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97bdc1ea2c7a6fd6366611611f6a221a858ff70b6ddda7c73893fac8ef8b4f3c","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275059,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[0,0],\"matrix\":[[-4,-2,-3],[1,4,-5]],\"u\":[1,1,-2],\"v\":[3,1,-2]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-4, -2, -3], [1, 4, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[0, 0], u=[1, 1, -2], v=[3, 1, -2]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 1, -2], [3, 1, -2], [0, 0], [-8, 32], [-8, 32], [0, 0], Yes)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 1, -2], [3, 1, -2], [0, 0], [-8, 32], [-8, 32], [0, 0], Yes)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0060","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4198d5aff2a17f472acb7996c591b00376c16daaa0499f0ad6290fdda4c25576","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275060,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[5,-1],\"matrix\":[[-3,-1,0],[2,3,0]],\"u\":[1,2,-1],\"v\":[4,1,-1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-3, -1, 0], [2, 3, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[5, -1], u=[1, 2, -1], v=[4, 1, -1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 2, -1], [4, 1, -1], [5, -1], [-13, 18], [-8, 17], [-5, 1], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 2, -1], [4, 1, -1], [5, -1], [-13, 18], [-8, 17], [-5, 1], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0061","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef64005b81d303883cf84a6d43c25fdae1d1445b91b5007b9a5cff8480e6370e","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275061,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[1,0],\"matrix\":[[-2,0,3],[-2,2,5]],\"u\":[1,-2,0],\"v\":[5,1,0]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-2, 0, 3], [-2, 2, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[1, 0], u=[1, -2, 0], v=[5, 1, 0]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -2, 0], [5, 1, 0], [1, 0], [-11, -14], [-10, -14], [-1, 0], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -2, 0], [5, 1, 0], [1, 0], [-11, -14], [-10, -14], [-1, 0], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0062","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65d807623b03ea56860c75274781969da8ed1bd2ddc39053a984fe5e96650526","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275062,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[2,1],\"matrix\":[[-1,1,-5],[-1,1,-3]],\"u\":[1,-1,1],\"v\":[-5,1,1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[-1, 1, -5], [-1, 1, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[2, 1], u=[1, -1, 1], v=[-5, 1, 1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -1, 1], [-5, 1, 1], [2, 1], [-4, -1], [-2, 0], [-2, -1], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -1, 1], [-5, 1, 1], [2, 1], [-4, -1], [-2, 0], [-2, -1], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0063","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e425b9ee0daffdbb98061e7b8c147ddc8c750b3d66e3e252e36c95e5e1a99607","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275063,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[0,0],\"matrix\":[[0,2,-2],[0,0,2]],\"u\":[1,0,2],\"v\":[-4,1,-2]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[0, 2, -2], [0, 0, 2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[0, 0], u=[1, 0, 2], v=[-4, 1, -2]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 0, 2], [-4, 1, -2], [0, 0], [2, 0], [2, 0], [0, 0], Yes)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 0, 2], [-4, 1, -2], [0, 0], [2, 0], [2, 0], [0, 0], Yes)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0064","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40b0e893003bde431fe9b22eec47f3e0931ab31b632139c7eddd9aca3dc5582c","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275064,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[4,3],\"matrix\":[[1,3,1],[1,-1,-6]],\"u\":[1,1,3],\"v\":[-3,1,-1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[1, 3, 1], [1, -1, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[4, 3], u=[1, 1, 3], v=[-3, 1, -1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 1, 3], [-3, 1, -1], [4, 3], [10, -13], [14, -10], [-4, -3], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 1, 3], [-3, 1, -1], [4, 3], [10, -13], [14, -10], [-4, -3], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0065","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de71c64e81b65a987e52c01b376934da1b8fa5e4379030d915eb8e9e78e79e4b","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275065,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[5,-3],\"matrix\":[[2,-3,4],[2,-2,-1]],\"u\":[1,2,-3],\"v\":[-2,1,0]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[2, -3, 4], [2, -2, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[5, -3], u=[1, 2, -3], v=[-2, 1, 0]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 2, -3], [-2, 1, 0], [5, -3], [-18, -8], [-13, -11], [-5, 3], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 2, -3], [-2, 1, 0], [5, -3], [-18, -8], [-13, -11], [-5, 3], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0066","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"389e0cc1f36a79065c0f85c47fb6cb8d902a7f4488e2d80fa8eff615e6111e9e","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275066,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[1,-2],\"matrix\":[[3,-2,-4],[-2,-3,4]],\"u\":[1,-2,-2],\"v\":[-1,1,1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[3, -2, -4], [-2, -3, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[1, -2], u=[1, -2, -2], v=[-1, 1, 1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -2, -2], [-1, 1, 1], [1, -2], [7, -3], [8, -5], [-1, 2], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -2, -2], [-1, 1, 1], [1, -2], [7, -3], [8, -5], [-1, 2], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0067","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1d8d015cc1f6656d1fd3b1e05c484ea4ce4dc8a5af3136775a6bc9cce867e12","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275067,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[0,0],\"matrix\":[[4,-1,-1],[-1,-4,-4]],\"u\":[1,-1,-1],\"v\":[0,1,-2]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[4, -1, -1], [-1, -4, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[0, 0], u=[1, -1, -1], v=[0, 1, -2]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -1, -1], [0, 1, -2], [0, 0], [7, 11], [7, 11], [0, 0], Yes)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -1, -1], [0, 1, -2], [0, 0], [7, 11], [7, 11], [0, 0], Yes)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0068","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"303f08b7ee04f7b8007abd8d0f4e095565c389e56f7b2c56e87aa78a9a88322d","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275068,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[3,0],\"matrix\":[[5,0,2],[0,-5,1]],\"u\":[1,0,0],\"v\":[1,1,-1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[5, 0, 2], [0, -5, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[3, 0], u=[1, 0, 0], v=[1, 1, -1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 0, 0], [1, 1, -1], [3, 0], [11, -6], [14, -6], [-3, 0], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 0, 0], [1, 1, -1], [3, 0], [11, -6], [14, -6], [-3, 0], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0069","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"781f862165288fd78b4cc4e24f53b17063e08b99b0a66d367843eb2f4b1bb72b","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275069,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[4,1],\"matrix\":[[6,1,5],[1,-6,6]],\"u\":[1,1,1],\"v\":[2,1,0]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[6, 1, 5], [1, -6, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[4, 1], u=[1, 1, 1], v=[2, 1, 0]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 1, 1], [2, 1, 0], [4, 1], [29, -2], [33, -1], [-4, -1], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 1, 1], [2, 1, 0], [4, 1], [29, -2], [33, -1], [-4, -1], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0070","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cce10c731ea0e3d86c5c0a594e4fb86502c1b39016c59cd3d63c85407a2af6a","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275070,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[5,2],\"matrix\":[[7,2,-3],[2,-7,-2]],\"u\":[1,2,2],\"v\":[3,1,1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[7, 2, -3], [2, -7, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[5, 2], u=[1, 2, 2], v=[3, 1, 1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 2, 2], [3, 1, 1], [5, 2], [30, -17], [35, -15], [-5, -2], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 2, 2], [3, 1, 1], [5, 2], [30, -17], [35, -15], [-5, -2], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0071","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8fdb0eac1e312d6f0cd84749489e87877da38bb5a63e8ab64c96066b32ea4c85","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275071,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[0,0],\"matrix\":[[8,3,0],[-2,-8,3]],\"u\":[1,-2,3],\"v\":[4,1,-2]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[8, 3, 0], [-2, -8, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[0, 0], u=[1, -2, 3], v=[4, 1, -2]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -2, 3], [4, 1, -2], [0, 0], [37, 1], [37, 1], [0, 0], Yes)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -2, 3], [4, 1, -2], [0, 0], [37, 1], [37, 1], [0, 0], Yes)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0072","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97993c3e0ea845e7b1057ff143d6e93165896201f3102d55d43e710d0025ef56","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275072,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[2,-3],\"matrix\":[[9,-3,3],[-1,-9,-5]],\"u\":[1,-1,-3],\"v\":[5,1,-1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[9, -3, 3], [-1, -9, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[2, -3], u=[1, -1, -3], v=[5, 1, -1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, -1, -3], [5, 1, -1], [2, -3], [44, 11], [46, 8], [-2, 3], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, -1, -3], [5, 1, -1], [2, -3], [44, 11], [46, 8], [-2, 3], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0073","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec8ad31233afcf5465e6e1bfab92b042e319d58ee9a438de00e7ed03427d682d","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275073,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[3,-2],\"matrix\":[[10,-2,-5],[0,-10,0]],\"u\":[1,0,-2],\"v\":[-5,1,0]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[10, -2, -5], [0, -10, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[3, -2], u=[1, 0, -2], v=[-5, 1, 0]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 0, -2], [-5, 1, 0], [3, -2], [-29, -12], [-26, -14], [-3, 2], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 0, -2], [-5, 1, 0], [3, -2], [-29, -12], [-26, -14], [-3, 2], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0074","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c221a76db16cfd59db6662a6b558097215e9f77f901b147c64ae1151b237aa37","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275074,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[4,-1],\"matrix\":[[11,-1,-2],[1,-11,5]],\"u\":[1,1,-1],\"v\":[-4,1,1]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[11, -1, -2], [1, -11, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[4, -1], u=[1, 1, -1], v=[-4, 1, 1]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 1, -1], [-4, 1, 1], [4, -1], [-31, -26], [-27, -27], [-4, 1], No)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 1, -1], [-4, 1, 1], [4, -1], [-31, -26], [-27, -27], [-4, 1], No)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0075","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b7381479cf4b77a1fc5dd480d795d71447637fd7ce82259b8500f10decc561e","family_id":"qf-mod27-linearmap-additivity-witness","seed":202610275075,"semantic_template":"qf-mod27-linearmap-additivity-witness/{\"bias\":[0,0],\"matrix\":[[12,0,1],[2,-12,-3]],\"u\":[1,2,0],\"v\":[-3,1,-2]}"},"fields":{"title":"Compare Both Sides of Additivity for the Supplied Affine Map","stem":"Use the complete rectangular matrix A=[[12, 0, 1], [2, -12, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. T(v)=A*v+bias, bias=[0, 0], u=[1, 2, 0], v=[-3, 1, -2]. Give (u,v,bias,T(u+v),T(u)+T(v),lhs−rhs,additive for this instance?).","answer":{"kind":"expression","value":"([1, 2, 0], [-3, 1, -2], [0, 0], [-26, -34], [-26, -34], [0, 0], Yes)","explanation":"The left side contains one bias and the right side contains two, so their difference is the negative bias. Retain that signed vector. No finite passing sample proves arbitrary black-box linearity. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is ([1, 2, 0], [-3, 1, -2], [0, 0], [-26, -34], [-26, -34], [0, 0], Yes)."},"skills":["Compare Both Sides of Additivity for the Supplied Affine Map"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0076","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72e137addba4f9cf588e69c34697cfdc16a6b0a0b89e2689e963f12d63c80b61","family_id":"qf-mod27-linearmap-combination-image","seed":202610275076,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-2,\"beta\":-3,\"matrix\":[[-12,-3,-5],[-2,12,-6]],\"u\":[1,-2,-3],\"v\":[-5,1,-2]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-12, -3, -5], [-2, 12, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -2, -3], v=[-5, 1, -2], alpha=-2, beta=-3; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-2, -3, [13, 1, 12], [-219, -86], [-219, -86], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-2, -3, [13, 1, 12], [-219, -86], [-219, -86], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0077","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a596ac94709722590330421c07da3a26a621b1d36d78687e57d166c2e9e2331c","family_id":"qf-mod27-linearmap-combination-image","seed":202610275077,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-1,\"beta\":-2,\"matrix\":[[-11,-2,-2],[-1,11,-1]],\"u\":[1,-1,-2],\"v\":[-4,1,-1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-11, -2, -2], [-1, 11, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -1, -2], v=[-4, 1, -1], alpha=-1, beta=-2; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-1, -2, [7, -1, 4], [-83, -22], [-83, -22], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-1, -2, [7, -1, 4], [-83, -22], [-83, -22], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0078","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"337240adb77081b56e97b65b4293cfc1dbe4dcdf0fd35977efa3d166688c1a2e","family_id":"qf-mod27-linearmap-combination-image","seed":202610275078,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":0,\"beta\":-1,\"matrix\":[[-10,-1,1],[0,10,4]],\"u\":[1,0,-1],\"v\":[-3,1,0]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-10, -1, 1], [0, 10, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 0, -1], v=[-3, 1, 0], alpha=0, beta=-1; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(0, -1, [3, -1, 0], [-29, -10], [-29, -10], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, -1, [3, -1, 0], [-29, -10], [-29, -10], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0079","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ad157efcc63420113f7b9acb494141c3ad88a16767524aa479b39ef5ea02362","family_id":"qf-mod27-linearmap-combination-image","seed":202610275079,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":1,\"beta\":0,\"matrix\":[[-9,0,4],[1,9,-4]],\"u\":[1,1,0],\"v\":[-2,1,1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-9, 0, 4], [1, 9, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 1, 0], v=[-2, 1, 1], alpha=1, beta=0; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(1, 0, [1, 1, 0], [-9, 10], [-9, 10], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 0, [1, 1, 0], [-9, 10], [-9, 10], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0080","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45fb20dcff681793b4acc866b6663d59a964687da072eff891d990c4045fbdb7","family_id":"qf-mod27-linearmap-combination-image","seed":202610275080,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":2,\"beta\":1,\"matrix\":[[-8,1,-4],[2,8,1]],\"u\":[1,2,1],\"v\":[-1,1,-2]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-8, 1, -4], [2, 8, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 2, 1], v=[-1, 1, -2], alpha=2, beta=1; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(2, 1, [1, 5, 0], [-3, 42], [-3, 42], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 1, [1, 5, 0], [-3, 42], [-3, 42], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0081","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd4eb26fd921f09d55681d5ea7228d8f3060bc3b05d438bda38403b78cc01030","family_id":"qf-mod27-linearmap-combination-image","seed":202610275081,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-2,\"beta\":2,\"matrix\":[[-7,2,-1],[-2,7,6]],\"u\":[1,-2,2],\"v\":[0,1,-1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-7, 2, -1], [-2, 7, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -2, 2], v=[0, 1, -1], alpha=-2, beta=2; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-2, 2, [-2, 6, -6], [32, 10], [32, 10], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-2, 2, [-2, 6, -6], [32, 10], [32, 10], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0082","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"379a118d35a446d58224683b4593dca446dc6a37aae53b7b2484e471bd2c7cbf","family_id":"qf-mod27-linearmap-combination-image","seed":202610275082,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-1,\"beta\":3,\"matrix\":[[-6,3,2],[-1,6,-2]],\"u\":[1,-1,3],\"v\":[1,1,0]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-6, 3, 2], [-1, 6, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -1, 3], v=[1, 1, 0], alpha=-1, beta=3; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-1, 3, [2, 4, -3], [-6, 28], [-6, 28], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-1, 3, [2, 4, -3], [-6, 28], [-6, 28], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0083","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b000111383b19a3d45538143e076da3b040cb0d6667863d81943f90eaec2838d","family_id":"qf-mod27-linearmap-combination-image","seed":202610275083,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":0,\"beta\":-3,\"matrix\":[[-5,-3,5],[0,5,3]],\"u\":[1,0,-3],\"v\":[2,1,1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-5, -3, 5], [0, 5, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 0, -3], v=[2, 1, 1], alpha=0, beta=-3; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(0, -3, [-6, -3, -3], [24, -24], [24, -24], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, -3, [-6, -3, -3], [24, -24], [24, -24], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0084","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bd7c61a1a7007ac9bf5c63e8e871c9a293956f7a314ec9ea5e0e7ae15243ef1","family_id":"qf-mod27-linearmap-combination-image","seed":202610275084,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":1,\"beta\":-2,\"matrix\":[[-4,-2,-3],[1,4,-5]],\"u\":[1,1,-2],\"v\":[3,1,-2]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-4, -2, -3], [1, 4, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 1, -2], v=[3, 1, -2], alpha=1, beta=-2; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(1, -2, [-5, -1, 2], [16, -19], [16, -19], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, -2, [-5, -1, 2], [16, -19], [16, -19], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0085","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79000054c9364f1839b9b8cd63c1bebc3eecde27aa341e28af5c71748a9c7384","family_id":"qf-mod27-linearmap-combination-image","seed":202610275085,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":2,\"beta\":-1,\"matrix\":[[-3,-1,0],[2,3,0]],\"u\":[1,2,-1],\"v\":[4,1,-1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-3, -1, 0], [2, 3, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 2, -1], v=[4, 1, -1], alpha=2, beta=-1; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(2, -1, [-2, 3, -1], [3, 5], [3, 5], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, -1, [-2, 3, -1], [3, 5], [3, 5], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0086","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"254dbb6c9b614814be2f8213a445b3d9ae89f57f2cd9ef94b712310fc0919edd","family_id":"qf-mod27-linearmap-combination-image","seed":202610275086,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-2,\"beta\":0,\"matrix\":[[-2,0,3],[-2,2,5]],\"u\":[1,-2,0],\"v\":[5,1,0]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-2, 0, 3], [-2, 2, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -2, 0], v=[5, 1, 0], alpha=-2, beta=0; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-2, 0, [-2, 4, 0], [4, 12], [4, 12], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-2, 0, [-2, 4, 0], [4, 12], [4, 12], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0087","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb9c899242db6d36d908560c709becdd70bd6cfad39a77cafc28d23f2b34c1be","family_id":"qf-mod27-linearmap-combination-image","seed":202610275087,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-1,\"beta\":1,\"matrix\":[[-1,1,-5],[-1,1,-3]],\"u\":[1,-1,1],\"v\":[-5,1,1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[-1, 1, -5], [-1, 1, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -1, 1], v=[-5, 1, 1], alpha=-1, beta=1; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-1, 1, [-6, 2, 0], [8, 8], [8, 8], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-1, 1, [-6, 2, 0], [8, 8], [8, 8], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0088","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a48efcdadd7a47e0810ccffe2c88c1606bb8ccc09af97e7866d990d85469589c","family_id":"qf-mod27-linearmap-combination-image","seed":202610275088,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":0,\"beta\":2,\"matrix\":[[0,2,-2],[0,0,2]],\"u\":[1,0,2],\"v\":[-4,1,-2]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[0, 2, -2], [0, 0, 2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 0, 2], v=[-4, 1, -2], alpha=0, beta=2; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(0, 2, [-8, 2, -4], [12, -8], [12, -8], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 2, [-8, 2, -4], [12, -8], [12, -8], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0089","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"930f6a9776a8da8145c0c6e9f8eca865680a4b193748f5754a35c00a6001d50c","family_id":"qf-mod27-linearmap-combination-image","seed":202610275089,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":1,\"beta\":3,\"matrix\":[[1,3,1],[1,-1,-6]],\"u\":[1,1,3],\"v\":[-3,1,-1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[1, 3, 1], [1, -1, -6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 1, 3], v=[-3, 1, -1], alpha=1, beta=3; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(1, 3, [-8, 4, 0], [4, -12], [4, -12], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 3, [-8, 4, 0], [4, -12], [4, -12], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0090","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82c8a5b07c503a78fa7039a1148d70ae5e1dfbbe1c3c82dc0b78d9de2921b814","family_id":"qf-mod27-linearmap-combination-image","seed":202610275090,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":2,\"beta\":-3,\"matrix\":[[2,-3,4],[2,-2,-1]],\"u\":[1,2,-3],\"v\":[-2,1,0]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[2, -3, 4], [2, -2, -1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 2, -3], v=[-2, 1, 0], alpha=2, beta=-3; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(2, -3, [8, 1, -6], [-11, 20], [-11, 20], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, -3, [8, 1, -6], [-11, 20], [-11, 20], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0091","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78bcd3ffeaf1b46a007ad12b08058369c787ab87db598ae53373e6db5a821e70","family_id":"qf-mod27-linearmap-combination-image","seed":202610275091,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-2,\"beta\":-2,\"matrix\":[[3,-2,-4],[-2,-3,4]],\"u\":[1,-2,-2],\"v\":[-1,1,1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[3, -2, -4], [-2, -3, 4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -2, -2], v=[-1, 1, 1], alpha=-2, beta=-2; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-2, -2, [0, 2, 2], [-12, 2], [-12, 2], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-2, -2, [0, 2, 2], [-12, 2], [-12, 2], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0092","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22ddcc3ed3e6325be52088c53b2a69b433b54edcb64deedf73f51b3645ba721d","family_id":"qf-mod27-linearmap-combination-image","seed":202610275092,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-1,\"beta\":-1,\"matrix\":[[4,-1,-1],[-1,-4,-4]],\"u\":[1,-1,-1],\"v\":[0,1,-2]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[4, -1, -1], [-1, -4, -4]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -1, -1], v=[0, 1, -2], alpha=-1, beta=-1; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-1, -1, [-1, 0, 3], [-7, -11], [-7, -11], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-1, -1, [-1, 0, 3], [-7, -11], [-7, -11], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0093","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca373e70ff5a6dd7920235f1d7b76c912f8e45d2bab65938ade9dfb55b6fa3b9","family_id":"qf-mod27-linearmap-combination-image","seed":202610275093,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":0,\"beta\":0,\"matrix\":[[5,0,2],[0,-5,1]],\"u\":[1,0,0],\"v\":[1,1,-1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[5, 0, 2], [0, -5, 1]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 0, 0], v=[1, 1, -1], alpha=0, beta=0; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(0, 0, [0, 0, 0], [0, 0], [0, 0], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, 0, [0, 0, 0], [0, 0], [0, 0], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0094","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee3ae70b3e4b8f1c4542ba58d7ecb2d86793010d17e0ea1dfb6c8fb658599b59","family_id":"qf-mod27-linearmap-combination-image","seed":202610275094,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":1,\"beta\":1,\"matrix\":[[6,1,5],[1,-6,6]],\"u\":[1,1,1],\"v\":[2,1,0]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[6, 1, 5], [1, -6, 6]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 1, 1], v=[2, 1, 0], alpha=1, beta=1; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(1, 1, [3, 2, 1], [25, -3], [25, -3], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, 1, [3, 2, 1], [25, -3], [25, -3], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0095","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4d3e7b0192b81ffb7c1e355aba34cfa57622f2bd87449aac46fac28a3237056","family_id":"qf-mod27-linearmap-combination-image","seed":202610275095,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":2,\"beta\":2,\"matrix\":[[7,2,-3],[2,-7,-2]],\"u\":[1,2,2],\"v\":[3,1,1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[7, 2, -3], [2, -7, -2]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 2, 2], v=[3, 1, 1], alpha=2, beta=2; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(2, 2, [8, 6, 6], [50, -38], [50, -38], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 2, [8, 6, 6], [50, -38], [50, -38], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0096","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b8e2aa7820189e99f974fc6e7db9ec77353a06681e2cf749972f2b2f7f4a638","family_id":"qf-mod27-linearmap-combination-image","seed":202610275096,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-2,\"beta\":3,\"matrix\":[[8,3,0],[-2,-8,3]],\"u\":[1,-2,3],\"v\":[4,1,-2]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[8, 3, 0], [-2, -8, 3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -2, 3], v=[4, 1, -2], alpha=-2, beta=3; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-2, 3, [10, 7, -12], [101, -112], [101, -112], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-2, 3, [10, 7, -12], [101, -112], [101, -112], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0097","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4eea069621866cfac6fc5a1fb1f50baadaed3b50f288bc6855d3c2e90a380ca0","family_id":"qf-mod27-linearmap-combination-image","seed":202610275097,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":-1,\"beta\":-3,\"matrix\":[[9,-3,3],[-1,-9,-5]],\"u\":[1,-1,-3],\"v\":[5,1,-1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[9, -3, 3], [-1, -9, -5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, -1, -3], v=[5, 1, -1], alpha=-1, beta=-3; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(-1, -3, [-16, -2, 6], [-120, 4], [-120, 4], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (-1, -3, [-16, -2, 6], [-120, 4], [-120, 4], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0098","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3bd5ac29ed2ecfe9eb757354dc0e9db224cdc727754ae5f0601769e734e3d32f","family_id":"qf-mod27-linearmap-combination-image","seed":202610275098,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":0,\"beta\":-2,\"matrix\":[[10,-2,-5],[0,-10,0]],\"u\":[1,0,-2],\"v\":[-5,1,0]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[10, -2, -5], [0, -10, 0]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 0, -2], v=[-5, 1, 0], alpha=0, beta=-2; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(0, -2, [10, -2, 0], [104, 20], [104, 20], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (0, -2, [10, -2, 0], [104, 20], [104, 20], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0099","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70d33cce6b065d7e21121b60229d79de6a370682582cf81c09a6c5ecbbefd34a","family_id":"qf-mod27-linearmap-combination-image","seed":202610275099,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":1,\"beta\":-1,\"matrix\":[[11,-1,-2],[1,-11,5]],\"u\":[1,1,-1],\"v\":[-4,1,1]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[11, -1, -2], [1, -11, 5]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 1, -1], v=[-4, 1, 1], alpha=1, beta=-1; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(1, -1, [5, 0, -2], [59, -5], [59, -5], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (1, -1, [5, 0, -2], [59, -5], [59, -5], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261027-r05-general-linear-map-0100","batch_id":"batch-20261014-advance-20261027-r05-general-linear-map-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ba41ea3d4b645e16d88ca25127af505a3cb751f9c500c7145e072903aa67049","family_id":"qf-mod27-linearmap-combination-image","seed":202610275100,"semantic_template":"qf-mod27-linearmap-combination-image/{\"alpha\":2,\"beta\":0,\"matrix\":[[12,0,1],[2,-12,-3]],\"u\":[1,2,0],\"v\":[-3,1,-2]}"},"fields":{"title":"The Matrix Formula Preserves a Whole Linear Combination","stem":"Use the complete rectangular matrix A=[[12, 0, 1], [2, -12, -3]] as a map from its column-coordinate space to its row-coordinate space, both over the reals. Matrix distributivity is a universal coordinate certificate. All vectors and any bias must have their respective input/output dimensions. Finite samples do not establish linearity of an arbitrary black box, or imply injectivity, surjectivity, boundedness or continuity beyond the stated claim. u=[1, 2, 0], v=[-3, 1, -2], alpha=2, beta=0; no bias is supplied. Give (alpha,beta,combined,lhs=A*combined,rhs=alpha*A*u+beta*A*v,equal?,matrix_formula_universal_certificate).","answer":{"kind":"expression","value":"(2, 0, [2, 4, 0], [24, -44], [24, -44], Yes, matrix_formula_universal_certificate)","explanation":"Use distributivity of each row dot product, then compute the two exact images in the target-coordinate space. The universal certificate comes from the formula, not just this instance. An exact finite or coordinate slice does not automatically establish a general infinite theorem. The exact result, retaining the requested order, is (2, 0, [2, 4, 0], [24, -44], [24, -44], Yes, matrix_formula_universal_certificate)."},"skills":["The Matrix Formula Preserves a Whole Linear Combination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}