{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261016-r05-system-linear-equations-0001","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d29fe783d92fc384c7769e60e2b5b89a1838f73ba969af184cf8ce977a5c43a","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165001,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[-1,-3],[1,3]],\"rows\":[[2,1,1],[1,-1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[2, 1, 1], [1, -1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[-1/-3, 1/3], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([0/1, 0/1], Yes)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0/1, 0/1], Yes)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0002","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc240bbb72901f697a71dffe71d06d17817c3ec78719d5d97c7641130eab2599","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165002,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,0],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 0], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([0/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0003","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d2e71f57f5a55afb818b8f3ceacebad0c2e6e34a8b8952ea8846fe7a01b7e53","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165003,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[0,0,0],[0,0,2]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[0, 0, 0], [0, 0, 2]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([0/1, -2/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0/1, -2/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0004","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc924a433644966e0566a999fbfbf9e7fc632e3a05a710e3b8ff33588bfa35e2","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165004,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,-9],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -9], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([9/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([9/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0005","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e5cb775e353002132685ead5c22dc532e93cb35921a6d940f38f68a4160944a","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165005,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,-8],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -8], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([8/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([8/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0006","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91af9d47ae93d7f23c75f70789e60206200a0813cf521656861bdb5b457bc64d","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165006,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,-7],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -7], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([7/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([7/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0007","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2c5ddca84a111f9ae43c3bf2d24bd9c397fc1453b39922dfb69db210089ee3d","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165007,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,-6],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -6], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([6/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([6/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0008","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5f9cc8c30d7a6e4066ba38bf2869b2d8ee5ac789b8f686eccd9cbf0146540f6","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165008,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,-5],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -5], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([5/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0009","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c45fca31effa01e6d7ba53ab77e121151fe477296c5338ecac1e166a35aa4863","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165009,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,-4],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -4], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([4/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([4/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0010","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96f499ed7c670f15ef4567fc2b1e6ef791bc59d320e8ad7dd3b69bac012b7b9a","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165010,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,-3],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -3], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([3/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([3/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0011","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c19365a832c8293c29a8bea0a8bb92a7320949918a19a8301908a101f3f8e87","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165011,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,-2],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -2], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([2/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0012","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca2e5271b1ee0b1b1d7e49f2859b2c6c875443ec106374eecf7b20990814e236","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165012,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,-1],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, -1], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([1/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0013","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5725285a180a62b054a23ed5f3890ee160a0ba5b6d518a6ce3f312a8dc54284","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165013,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[2,1]],\"rows\":[[1,0,0],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 0], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 2/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([0/1, 2/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0/1, 2/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0014","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb2393030ab895fb2c7b460344446111034d20a6e10c5b53b01edac4bcf880e3","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165014,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,1],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 1], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-1/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0015","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"407c71f43d5850fe15ac78cbd02cffd36980279e31cd4a9042bf3d1b3733e4b5","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165015,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,2],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 2], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-2/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-2/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0016","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d0af4705160eb1ed4acd76eb2c0b6fb41b621280e726dffd1b55a3b080b4ed3","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165016,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,3],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 3], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-3/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-3/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0017","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3afd73c6a34b6a1717083cea9bb46ce1ed8df817fab959e9f9363724c5b28b7","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165017,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,4],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 4], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-4/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-4/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0018","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf4b4db9bc74c19d89e4ec8ddbda51ad9052ae0bf5876a1c1415990b5a129efd","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165018,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,5],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 5], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-5/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-5/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0019","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdc85e309b472f00bc395c1fa23ee0a9de14dcaf9924f9268605191f7dfaf81d","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165019,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,6],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 6], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-6/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-6/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0020","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db6e56a27b00ba526a1021e2692a16161075b2ebd87bae98de7eaa67abba08b8","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165020,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,7],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 7], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-7/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-7/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0021","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c4137c301e21bf2df6d45768bdb8e0b047f4421ea46e3761e13dfd930545ada","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165021,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,8],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 8], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-8/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-8/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0022","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43f06b7569da3a465847ad885863549cdd0359b92f5b3bad4ebf7f046f66760e","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165022,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,9],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 9], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-9/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-9/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0023","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a5feaf7f871f529961104190ca04f05461bb9f65daf6e97280efa593ad4cc77","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165023,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,10],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 10], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-10/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-10/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0024","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"918c8b665db4677141ff57d001cd691b5517376d5f591c5fb359b76062a67ae5","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165024,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[1,1]],\"rows\":[[1,0,11],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 11], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 1/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-11/1, 1/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-11/1, 1/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0025","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17fc6f41e8ac068ca986520514ce53f1eb700273b543687a3e86630a79e5a1e4","family_id":"qf-density16-system-candidate-two-residuals","seed":202610165025,"semantic_template":"qf-density16-system-candidate-two-residuals/{\"candidate\":[[0,1],[0,1]],\"rows\":[[1,0,12],[0,1,0]]}"},"fields":{"title":"Two Rational Substitution Residuals for a System","stem":"Over R, rows=[[1, 0, 12], [0, 1, 0]], where each [a,b,c] means a*x+b*y=c. Ordered candidate (x,y)=[0/1, 0/1], with nonzero representation denominators. Give (two exact rational a*x+b*y−c residuals, are both zero?).","answer":{"kind":"expression","value":"([-12/1, 0/1], No)","explanation":"Normalize valid candidate fractions, including negative denominators, then substitute into both equations; passing one equation alone is insufficient. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-12/1, 0/1], No)."},"skills":["Two Rational Substitution Residuals for a System"],"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-20261012-advance-20261016-r05-system-linear-equations-0026","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20a64cb8455b74fb4ff58964fbd216217eeb6f9a349530d4f550a76bfe693468","family_id":"qf-density16-system-classify-solve","seed":202610165026,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[0,0,0],[0,0,0]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[0, 0, 0], [0, 0, 0]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(0, 0, 0, infinitely-many-solutions, not-unique)","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 0, 0, infinitely-many-solutions, not-unique)."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0027","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c9a6d3943c7e876ba82a1bd9632f66c176083e3c28acbcec73c37061a36f0b7","family_id":"qf-density16-system-classify-solve","seed":202610165027,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[0,0,1],[0,0,0]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[0, 0, 1], [0, 0, 0]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(0, 0, 1, no-solution, not-applicable)","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 0, 1, no-solution, not-applicable)."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0028","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc81bf649f03e50b075a216ce3ab174ea4652f8634ae56f383c2fb51f198b503","family_id":"qf-density16-system-classify-solve","seed":202610165028,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,1],[2,2,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 1], [2, 2, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(0, 1, 1, infinitely-many-solutions, not-unique)","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 1, infinitely-many-solutions, not-unique)."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0029","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7bc6130e11c8b07836648d0b38ceb38a44f59da022e1e90016397a3cc69989a6","family_id":"qf-density16-system-classify-solve","seed":202610165029,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,1],[2,2,3]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 1], [2, 2, 3]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(0, 1, 2, no-solution, not-applicable)","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (0, 1, 2, no-solution, not-applicable)."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0030","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c0eea999004f6ad51ac4ad197a26b4167d77ef20c241174e8a89bca06ed2eaa","family_id":"qf-density16-system-classify-solve","seed":202610165030,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[2,1,1],[1,-1,0]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[2, 1, 1], [1, -1, 0]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(-3, 2, 2, unique-solution, [1/3, 1/3])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (-3, 2, 2, unique-solution, [1/3, 1/3])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0031","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aeaad9f928ca38ee455381af51d4fb4eb5321feb0f2afa168e5a60b12c8116bf","family_id":"qf-density16-system-classify-solve","seed":202610165031,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,-7],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, -7], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-9/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-9/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0032","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a73135eeb21c80ca0c20b4cf2fc4dea432631b1e413eb399724f8c1e922d6920","family_id":"qf-density16-system-classify-solve","seed":202610165032,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,-6],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, -6], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-8/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-8/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0033","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30373cc1abd91899cf22d8c0bb2530058f057de2ab574645f494ac20a8b99f52","family_id":"qf-density16-system-classify-solve","seed":202610165033,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,-5],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, -5], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-7/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-7/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0034","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"899607b2bf33eda25e561076936a43d77c55c7c44caf9e6cb59cffb824d61253","family_id":"qf-density16-system-classify-solve","seed":202610165034,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,-4],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, -4], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-6/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-6/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0035","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"71e839ef6220ee74f72d16da2a22076b00215a96537e5e5b0f22ecc871765772","family_id":"qf-density16-system-classify-solve","seed":202610165035,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,-3],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, -3], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-5/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-5/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0036","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe7353d86f243b1195aaee8872f50f3fd29aa1fea7f06734c9b6664623255763","family_id":"qf-density16-system-classify-solve","seed":202610165036,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,-2],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, -2], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-4/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-4/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0037","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae2d838dffb7426a6b0a6757c9cb61d7ca2d508223081d483b8cae39c13d2614","family_id":"qf-density16-system-classify-solve","seed":202610165037,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,-1],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, -1], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-3/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-3/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0038","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc159c8d2b36480b49455cece9f399fe088f459769457869d0dd3420b551cdfc","family_id":"qf-density16-system-classify-solve","seed":202610165038,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,0],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 0], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-2/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-2/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0039","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fcfd3c49a5b1dd1e64e93f1245df250e3f527028a1f5e70b4c37aa64bc391d91","family_id":"qf-density16-system-classify-solve","seed":202610165039,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,1],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 1], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [-1/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [-1/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0040","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39d32894e6657c5b778f4fc8f647814c37542c0d4034d5869bfd47ea7d113a2d","family_id":"qf-density16-system-classify-solve","seed":202610165040,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,2],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 2], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [0/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [0/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0041","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"054191e8980ae3431877c1dff81d33faa371530bd005948ba4c0c1d34440f1f6","family_id":"qf-density16-system-classify-solve","seed":202610165041,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,3],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 3], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [1/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [1/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0042","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf332d7eeac707b3cf9f51ea5d0eb07a6583970366602365ef95a0a739ede5dd","family_id":"qf-density16-system-classify-solve","seed":202610165042,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,4],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 4], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [2/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [2/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0043","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"039b988a83788c7d17aad59325a1bf40b3985a7ba09f8868beab902db02f09dc","family_id":"qf-density16-system-classify-solve","seed":202610165043,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,5],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 5], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [3/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [3/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0044","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4abcf65fef13a1e77839cae6834eed65e19a67f5319dbb8b90f44ad29546f81","family_id":"qf-density16-system-classify-solve","seed":202610165044,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,6],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 6], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [4/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [4/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0045","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48d5aaf12549b47b635d5a64910fda0a4f586fd0556852ac1209ac93bf97c901","family_id":"qf-density16-system-classify-solve","seed":202610165045,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,7],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 7], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [5/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [5/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0046","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8c9add01157608d57aced2cc8294d66fbb9159d8740d6a87d6626b3017f01e0","family_id":"qf-density16-system-classify-solve","seed":202610165046,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,8],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 8], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [6/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [6/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0047","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a91f6b9a4c8e9f232f1c94c9f4dccc49aaa002ffd652bf701492330039962e7","family_id":"qf-density16-system-classify-solve","seed":202610165047,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,9],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 9], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [7/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [7/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0048","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8af763d30b46de2b7df91ad598289218ca5751f4766eae85d7e639f5e9dd20df","family_id":"qf-density16-system-classify-solve","seed":202610165048,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,10],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 10], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [8/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [8/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0049","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0aa9dde1f0949f1eaf577a17a3a5c491b880ae3e95301561e2741b44926471a","family_id":"qf-density16-system-classify-solve","seed":202610165049,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,11],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 11], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [9/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [9/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0050","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0973e3fb814cf1c8a6366602fce65307636dce6593bfa079e1b0e55015c817b3","family_id":"qf-density16-system-classify-solve","seed":202610165050,"semantic_template":"qf-density16-system-classify-solve/{\"rows\":[[1,1,12],[0,1,2]]}"},"fields":{"title":"Signed Determinant, Both Ranks, and System Class","stem":"Over R, rows=[[1, 1, 12], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. Give [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique] as an ordered tuple. A zero determinant alone does not decide inconsistency.","answer":{"kind":"expression","value":"(1, 2, 2, unique-solution, [10/1, 2/1])","explanation":"Compare coefficient and augmented ranks using exact rational arithmetic: unequal ranks mean no solution, equal rank below 2 means infinitely many solutions, and rank 2 yields the ordered unique rational solution. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is (1, 2, 2, unique-solution, [10/1, 2/1])."},"skills":["Signed Determinant, Both Ranks, and System Class"],"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-20261012-advance-20261016-r05-system-linear-equations-0051","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"432d338c9944cd36eb6f14fbafebc48a2c06b09e4b30f5b16286892638ca9c77","family_id":"qf-density16-system-eliminate-row","seed":202610165051,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[0,0,0],[0,0,0]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[0, 0, 0], [0, 0, 0]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[0, 0, 0], [0, 0, 0]], [0, 0, 0, infinitely-many-solutions, not-unique], [0, 0, 0, infinitely-many-solutions, not-unique], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0, 0], [0, 0, 0]], [0, 0, 0, infinitely-many-solutions, not-unique], [0, 0, 0, infinitely-many-solutions, not-unique], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0052","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"616e667230e4f8a745200f3989b6beb60ac28721eccc09651007a8860af65250","family_id":"qf-density16-system-eliminate-row","seed":202610165052,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,1,1],[2,2,2]],\"t\":-2}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 1, 1], [2, 2, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 1, 1], [0, 0, 0]], [0, 1, 1, infinitely-many-solutions, not-unique], [0, 1, 1, infinitely-many-solutions, not-unique], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 1, 1], [0, 0, 0]], [0, 1, 1, infinitely-many-solutions, not-unique], [0, 1, 1, infinitely-many-solutions, not-unique], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0053","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a17c0523942a3677e07cf24e644c6a499c109e75288e14f5500ff07e66d672eb","family_id":"qf-density16-system-eliminate-row","seed":202610165053,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,1,1],[2,2,3]],\"t\":-2}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 1, 1], [2, 2, 3]], where each [a,b,c] means a*x+b*y=c. t=-2; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 1, 1], [0, 0, 1]], [0, 1, 2, no-solution, not-applicable], [0, 1, 2, no-solution, not-applicable], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 1, 1], [0, 0, 1]], [0, 1, 2, no-solution, not-applicable], [0, 1, 2, no-solution, not-applicable], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0054","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ea73429cb89cbf94a82872b2f302a6df85c1d7be3a99a0566109b4bce198bb8","family_id":"qf-density16-system-eliminate-row","seed":202610165054,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[2,1,1],[1,-1,0]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[2, 1, 1], [1, -1, 0]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[2, 1, 1], [-1, -2, -1]], [-3, 2, 2, unique-solution, [1/3, 1/3]], [-3, 2, 2, unique-solution, [1/3, 1/3]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[2, 1, 1], [-1, -2, -1]], [-3, 2, 2, unique-solution, [1/3, 1/3]], [-3, 2, 2, unique-solution, [1/3, 1/3]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0055","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"379c84e7151c229c623b4e043641ddf33a9aca317b1f369b603f289a002b047a","family_id":"qf-density16-system-eliminate-row","seed":202610165055,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-8],[0,1,2]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -8], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -8], [0, 1, 2]], [1, 2, 2, unique-solution, [-8/1, 2/1]], [1, 2, 2, unique-solution, [-8/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -8], [0, 1, 2]], [1, 2, 2, unique-solution, [-8/1, 2/1]], [1, 2, 2, unique-solution, [-8/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0056","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a57f6d51d3b1a0462e0f464f639656f4d2b856abe7487e1e73e6a3e465950c17","family_id":"qf-density16-system-eliminate-row","seed":202610165056,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-7],[0,1,2]],\"t\":1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -7], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -7], [1, 1, -5]], [1, 2, 2, unique-solution, [-7/1, 2/1]], [1, 2, 2, unique-solution, [-7/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -7], [1, 1, -5]], [1, 2, 2, unique-solution, [-7/1, 2/1]], [1, 2, 2, unique-solution, [-7/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0057","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d18793d379941dcfda35098a11611695840e8f75e675be583fb703dc962791dd","family_id":"qf-density16-system-eliminate-row","seed":202610165057,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-6],[0,1,2]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -6], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -6], [-1, 1, 8]], [1, 2, 2, unique-solution, [-6/1, 2/1]], [1, 2, 2, unique-solution, [-6/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -6], [-1, 1, 8]], [1, 2, 2, unique-solution, [-6/1, 2/1]], [1, 2, 2, unique-solution, [-6/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0058","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a49576b4eed4c97df0c3f6ad4600e5d19727b9af77c00639266550c087922814","family_id":"qf-density16-system-eliminate-row","seed":202610165058,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-5],[0,1,2]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -5], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -5], [0, 1, 2]], [1, 2, 2, unique-solution, [-5/1, 2/1]], [1, 2, 2, unique-solution, [-5/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -5], [0, 1, 2]], [1, 2, 2, unique-solution, [-5/1, 2/1]], [1, 2, 2, unique-solution, [-5/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0059","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f57192fc4d93c7f5715b6d936916d97ac71097d6bfc177b1c1e8226fae1e9c8","family_id":"qf-density16-system-eliminate-row","seed":202610165059,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-4],[0,1,2]],\"t\":1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -4], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -4], [1, 1, -2]], [1, 2, 2, unique-solution, [-4/1, 2/1]], [1, 2, 2, unique-solution, [-4/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -4], [1, 1, -2]], [1, 2, 2, unique-solution, [-4/1, 2/1]], [1, 2, 2, unique-solution, [-4/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0060","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f89a019083a219c89e7813acc7283c9ef80165072a739b0f709ce3ab9ae91c77","family_id":"qf-density16-system-eliminate-row","seed":202610165060,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-3],[0,1,2]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -3], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -3], [-1, 1, 5]], [1, 2, 2, unique-solution, [-3/1, 2/1]], [1, 2, 2, unique-solution, [-3/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -3], [-1, 1, 5]], [1, 2, 2, unique-solution, [-3/1, 2/1]], [1, 2, 2, unique-solution, [-3/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0061","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90698d9f5783d6a66c15c77138e64e2802cb19f8f23d73124db2b879e6b8b6ff","family_id":"qf-density16-system-eliminate-row","seed":202610165061,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-2],[0,1,2]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -2], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -2], [0, 1, 2]], [1, 2, 2, unique-solution, [-2/1, 2/1]], [1, 2, 2, unique-solution, [-2/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -2], [0, 1, 2]], [1, 2, 2, unique-solution, [-2/1, 2/1]], [1, 2, 2, unique-solution, [-2/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0062","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93c3ec0f79982a4c7fcbe5c646a414e6261cef63e0b246117587e142548c90c0","family_id":"qf-density16-system-eliminate-row","seed":202610165062,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,-1],[0,1,2]],\"t\":1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, -1], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, -1], [1, 1, 1]], [1, 2, 2, unique-solution, [-1/1, 2/1]], [1, 2, 2, unique-solution, [-1/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -1], [1, 1, 1]], [1, 2, 2, unique-solution, [-1/1, 2/1]], [1, 2, 2, unique-solution, [-1/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0063","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0a9c808b7a70e08e7d2b1e33be2fe3a33ff95fd13edaabe7c73fc24417ec7d8","family_id":"qf-density16-system-eliminate-row","seed":202610165063,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,0],[0,1,2]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 0], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 0], [-1, 1, 2]], [1, 2, 2, unique-solution, [0/1, 2/1]], [1, 2, 2, unique-solution, [0/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 0], [-1, 1, 2]], [1, 2, 2, unique-solution, [0/1, 2/1]], [1, 2, 2, unique-solution, [0/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0064","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8fdbb664cf45bc656b6b1e0edd5c6941ad92da9805edfa86a221aa059cd5f8b","family_id":"qf-density16-system-eliminate-row","seed":202610165064,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,1],[0,1,2]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 1], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 1], [0, 1, 2]], [1, 2, 2, unique-solution, [1/1, 2/1]], [1, 2, 2, unique-solution, [1/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 1], [0, 1, 2]], [1, 2, 2, unique-solution, [1/1, 2/1]], [1, 2, 2, unique-solution, [1/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0065","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e3e0245a2d84455345659ef594ce2be9ffe0d88070ce720eef439eca1db35db","family_id":"qf-density16-system-eliminate-row","seed":202610165065,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,2],[0,1,2]],\"t\":1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 2], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 2], [1, 1, 4]], [1, 2, 2, unique-solution, [2/1, 2/1]], [1, 2, 2, unique-solution, [2/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 2], [1, 1, 4]], [1, 2, 2, unique-solution, [2/1, 2/1]], [1, 2, 2, unique-solution, [2/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0066","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06459bddb648093db1c003c05bb4beef9005ce4746b17bbfdc1f10ee0bceb993","family_id":"qf-density16-system-eliminate-row","seed":202610165066,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,3],[0,1,2]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 3], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 3], [-1, 1, -1]], [1, 2, 2, unique-solution, [3/1, 2/1]], [1, 2, 2, unique-solution, [3/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 3], [-1, 1, -1]], [1, 2, 2, unique-solution, [3/1, 2/1]], [1, 2, 2, unique-solution, [3/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0067","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"964445e1c62b40efc1fe489cade74a0e10558fc87b5c88c431b9365ed6da0a5c","family_id":"qf-density16-system-eliminate-row","seed":202610165067,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,4],[0,1,2]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 4], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 4], [0, 1, 2]], [1, 2, 2, unique-solution, [4/1, 2/1]], [1, 2, 2, unique-solution, [4/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 4], [0, 1, 2]], [1, 2, 2, unique-solution, [4/1, 2/1]], [1, 2, 2, unique-solution, [4/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0068","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd6e4566c2734034bfcfbf2420c833b0ac3f49bf46e3195f467ddfa9ce09bf90","family_id":"qf-density16-system-eliminate-row","seed":202610165068,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,5],[0,1,2]],\"t\":1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 5], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 5], [1, 1, 7]], [1, 2, 2, unique-solution, [5/1, 2/1]], [1, 2, 2, unique-solution, [5/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 5], [1, 1, 7]], [1, 2, 2, unique-solution, [5/1, 2/1]], [1, 2, 2, unique-solution, [5/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0069","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4702bf59a8e374d5cfa3754880a1fe6d6b091885bcb0afc5a311d9de76b91cad","family_id":"qf-density16-system-eliminate-row","seed":202610165069,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,6],[0,1,2]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 6], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 6], [-1, 1, -4]], [1, 2, 2, unique-solution, [6/1, 2/1]], [1, 2, 2, unique-solution, [6/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 6], [-1, 1, -4]], [1, 2, 2, unique-solution, [6/1, 2/1]], [1, 2, 2, unique-solution, [6/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0070","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61ff38c6c102a9b489f02d2282b5d7ed7ef2db6d2f439f34232ffea6e85d8d8a","family_id":"qf-density16-system-eliminate-row","seed":202610165070,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,7],[0,1,2]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 7], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 7], [0, 1, 2]], [1, 2, 2, unique-solution, [7/1, 2/1]], [1, 2, 2, unique-solution, [7/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 7], [0, 1, 2]], [1, 2, 2, unique-solution, [7/1, 2/1]], [1, 2, 2, unique-solution, [7/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0071","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93d9eda96c28422de3bab4ea8e8e9c8838e6fe9facac98690f9b3c90e827a5cb","family_id":"qf-density16-system-eliminate-row","seed":202610165071,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,8],[0,1,2]],\"t\":1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 8], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 8], [1, 1, 10]], [1, 2, 2, unique-solution, [8/1, 2/1]], [1, 2, 2, unique-solution, [8/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 8], [1, 1, 10]], [1, 2, 2, unique-solution, [8/1, 2/1]], [1, 2, 2, unique-solution, [8/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0072","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16c6effe3a59d736e5a9c877ec439c2c0e66788d01df55b834c84d112c61b9e9","family_id":"qf-density16-system-eliminate-row","seed":202610165072,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,9],[0,1,2]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 9], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 9], [-1, 1, -7]], [1, 2, 2, unique-solution, [9/1, 2/1]], [1, 2, 2, unique-solution, [9/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 9], [-1, 1, -7]], [1, 2, 2, unique-solution, [9/1, 2/1]], [1, 2, 2, unique-solution, [9/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0073","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9d552685fde7d951d7f0ce1468d8afc92e8b7f6b65055d7d52a5528ecd0aaca","family_id":"qf-density16-system-eliminate-row","seed":202610165073,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,10],[0,1,2]],\"t\":0}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 10], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 10], [0, 1, 2]], [1, 2, 2, unique-solution, [10/1, 2/1]], [1, 2, 2, unique-solution, [10/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 10], [0, 1, 2]], [1, 2, 2, unique-solution, [10/1, 2/1]], [1, 2, 2, unique-solution, [10/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0074","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"573f42001645091cb2191a89f503b8fe6e16a8ee97f6f71846daeccf76910b0a","family_id":"qf-density16-system-eliminate-row","seed":202610165074,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,11],[0,1,2]],\"t\":1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 11], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 11], [1, 1, 13]], [1, 2, 2, unique-solution, [11/1, 2/1]], [1, 2, 2, unique-solution, [11/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 11], [1, 1, 13]], [1, 2, 2, unique-solution, [11/1, 2/1]], [1, 2, 2, unique-solution, [11/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0075","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b38c19331cb5acc89df2942f1dceeb5341d75872199e5c2b0b1fd2fcbe8aa6f","family_id":"qf-density16-system-eliminate-row","seed":202610165075,"semantic_template":"qf-density16-system-eliminate-row/{\"rows\":[[1,0,12],[0,1,2]],\"t\":-1}"},"fields":{"title":"Reversible Row Addition and Before/After Classification","stem":"Over R, rows=[[1, 0, 12], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-1; replace only row2 by row2+t*row1, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, invertible-row-operation eligibility).","answer":{"kind":"expression","value":"([[1, 0, 12], [-1, 1, -10]], [1, 2, 2, unique-solution, [12/1, 2/1]], [1, 2, 2, unique-solution, [12/1, 2/1]], Yes)","explanation":"Add the complete rows and recompute each classification; adding −t times the first row reverses this operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 12], [-1, 1, -10]], [1, 2, 2, unique-solution, [12/1, 2/1]], [1, 2, 2, unique-solution, [12/1, 2/1]], Yes)."},"skills":["Reversible Row Addition and Before/After Classification"],"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-20261012-advance-20261016-r05-system-linear-equations-0076","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72ee81c137586078ea2b35b32cbb83f903f8849a06974e4e6dbbba4e0b00147b","family_id":"qf-density16-system-row-scaling-solution","seed":202610165076,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[0,0,0],[0,0,0]],\"t\":0}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[0, 0, 0], [0, 0, 0]], where each [a,b,c] means a*x+b*y=c. t=0; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[0, 0, 0], [0, 0, 0]], [0, 0, 0, infinitely-many-solutions, not-unique], [0, 0, 0, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0, 0], [0, 0, 0]], [0, 0, 0, infinitely-many-solutions, not-unique], [0, 0, 0, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0077","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34889f190633562fa8a83bf0a5706df377a29e37896d331a25745dc58ae3d3cd","family_id":"qf-density16-system-row-scaling-solution","seed":202610165077,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,1,1],[2,2,2]],\"t\":0}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 1, 1], [2, 2, 2]], where each [a,b,c] means a*x+b*y=c. t=0; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[0, 0, 0], [2, 2, 2]], [0, 1, 1, infinitely-many-solutions, not-unique], [0, 1, 1, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0, 0], [2, 2, 2]], [0, 1, 1, infinitely-many-solutions, not-unique], [0, 1, 1, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0078","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e4b2609fcb4e8685b3136cc7174bf27d9f984a43e0a6cac6a4fb434549f2262","family_id":"qf-density16-system-row-scaling-solution","seed":202610165078,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,1],[0,1,2]],\"t\":0}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 1], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=0; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[0, 0, 0], [0, 1, 2]], [1, 2, 2, unique-solution, [1/1, 2/1]], [0, 1, 1, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0, 0], [0, 1, 2]], [1, 2, 2, unique-solution, [1/1, 2/1]], [0, 1, 1, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0079","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce4619cfd33084c32ea6fbd415d409596a850ab120630819c3fc2bbbb0f7d038","family_id":"qf-density16-system-row-scaling-solution","seed":202610165079,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,1,1],[2,2,3]],\"t\":0}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 1, 1], [2, 2, 3]], where each [a,b,c] means a*x+b*y=c. t=0; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[0, 0, 0], [2, 2, 3]], [0, 1, 2, no-solution, not-applicable], [0, 1, 1, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0, 0], [2, 2, 3]], [0, 1, 2, no-solution, not-applicable], [0, 1, 1, infinitely-many-solutions, not-unique], No, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0080","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6e8bd95dc6ebf4809c1e6316e87903b915aedd81ca4cf8a079ce30ca7266d4e","family_id":"qf-density16-system-row-scaling-solution","seed":202610165080,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-8],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -8], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, -8], [0, 1, 2]], [1, 2, 2, unique-solution, [-8/1, 2/1]], [1, 2, 2, unique-solution, [-8/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -8], [0, 1, 2]], [1, 2, 2, unique-solution, [-8/1, 2/1]], [1, 2, 2, unique-solution, [-8/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0081","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3878380e350faf0cccb1006db2e12588730069a400e28d0e7e124b0f581fe90","family_id":"qf-density16-system-row-scaling-solution","seed":202610165081,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-7],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -7], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, 14], [0, 1, 2]], [1, 2, 2, unique-solution, [-7/1, 2/1]], [-2, 2, 2, unique-solution, [-7/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, 14], [0, 1, 2]], [1, 2, 2, unique-solution, [-7/1, 2/1]], [-2, 2, 2, unique-solution, [-7/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0082","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c651fdd4c17a8a23e67d5e95f5e9af265045f7de6e47fa2adafefd34a288d92","family_id":"qf-density16-system-row-scaling-solution","seed":202610165082,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-6],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -6], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, -6], [0, 1, 2]], [1, 2, 2, unique-solution, [-6/1, 2/1]], [1, 2, 2, unique-solution, [-6/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -6], [0, 1, 2]], [1, 2, 2, unique-solution, [-6/1, 2/1]], [1, 2, 2, unique-solution, [-6/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0083","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d94c891831db6dc10029ffa69e5a15160e409e88ea33a9a684d2404bdae2bcf8","family_id":"qf-density16-system-row-scaling-solution","seed":202610165083,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-5],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -5], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, 10], [0, 1, 2]], [1, 2, 2, unique-solution, [-5/1, 2/1]], [-2, 2, 2, unique-solution, [-5/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, 10], [0, 1, 2]], [1, 2, 2, unique-solution, [-5/1, 2/1]], [-2, 2, 2, unique-solution, [-5/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0084","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3253d0eca950f22a254515b694835412f08cd24217acc31b781e814fd1be3254","family_id":"qf-density16-system-row-scaling-solution","seed":202610165084,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-4],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -4], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, -4], [0, 1, 2]], [1, 2, 2, unique-solution, [-4/1, 2/1]], [1, 2, 2, unique-solution, [-4/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -4], [0, 1, 2]], [1, 2, 2, unique-solution, [-4/1, 2/1]], [1, 2, 2, unique-solution, [-4/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0085","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6c6b7c4307582a066ce9e231dfc96dd2b2509518de4397c100007068e787b3d","family_id":"qf-density16-system-row-scaling-solution","seed":202610165085,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-3],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -3], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, 6], [0, 1, 2]], [1, 2, 2, unique-solution, [-3/1, 2/1]], [-2, 2, 2, unique-solution, [-3/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, 6], [0, 1, 2]], [1, 2, 2, unique-solution, [-3/1, 2/1]], [-2, 2, 2, unique-solution, [-3/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0086","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f98e715c8ff9aee331a4c3abfde9e732a99f601945448240fa81d2f14cbef9f2","family_id":"qf-density16-system-row-scaling-solution","seed":202610165086,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-2],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -2], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, -2], [0, 1, 2]], [1, 2, 2, unique-solution, [-2/1, 2/1]], [1, 2, 2, unique-solution, [-2/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, -2], [0, 1, 2]], [1, 2, 2, unique-solution, [-2/1, 2/1]], [1, 2, 2, unique-solution, [-2/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0087","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b02041795a7c19f8bf77b22e27617aad02a8cd49e977a6cd2b38832581e1b491","family_id":"qf-density16-system-row-scaling-solution","seed":202610165087,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,-1],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, -1], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, 2], [0, 1, 2]], [1, 2, 2, unique-solution, [-1/1, 2/1]], [-2, 2, 2, unique-solution, [-1/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, 2], [0, 1, 2]], [1, 2, 2, unique-solution, [-1/1, 2/1]], [-2, 2, 2, unique-solution, [-1/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0088","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1537d736fa9f801779c58b56d5a1eaa0c8509d944d86ed04e0c22f899357da37","family_id":"qf-density16-system-row-scaling-solution","seed":202610165088,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,0],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 0], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, 0], [0, 1, 2]], [1, 2, 2, unique-solution, [0/1, 2/1]], [1, 2, 2, unique-solution, [0/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 0], [0, 1, 2]], [1, 2, 2, unique-solution, [0/1, 2/1]], [1, 2, 2, unique-solution, [0/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0089","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"174830d514cf0b0e6ffe2a2e96680a4da3e255471b7e269b98b7bebdc97c8d52","family_id":"qf-density16-system-row-scaling-solution","seed":202610165089,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,1],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 1], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, -2], [0, 1, 2]], [1, 2, 2, unique-solution, [1/1, 2/1]], [-2, 2, 2, unique-solution, [1/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, -2], [0, 1, 2]], [1, 2, 2, unique-solution, [1/1, 2/1]], [-2, 2, 2, unique-solution, [1/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0090","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd64a1e405d16556cae5d035059dbd59b7fb587679419d1d28b8f93787af9ca8","family_id":"qf-density16-system-row-scaling-solution","seed":202610165090,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,2],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 2], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, 2], [0, 1, 2]], [1, 2, 2, unique-solution, [2/1, 2/1]], [1, 2, 2, unique-solution, [2/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 2], [0, 1, 2]], [1, 2, 2, unique-solution, [2/1, 2/1]], [1, 2, 2, unique-solution, [2/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0091","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a7d16315da79872e17a32a98e54cd5fc2a62d0c9fe002d66240681b5bbf9363","family_id":"qf-density16-system-row-scaling-solution","seed":202610165091,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,3],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 3], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, -6], [0, 1, 2]], [1, 2, 2, unique-solution, [3/1, 2/1]], [-2, 2, 2, unique-solution, [3/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, -6], [0, 1, 2]], [1, 2, 2, unique-solution, [3/1, 2/1]], [-2, 2, 2, unique-solution, [3/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0092","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb5463a7e894b83b47404dfca7dcabb853cb8f7f19156915c97c08f20beae94c","family_id":"qf-density16-system-row-scaling-solution","seed":202610165092,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,4],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 4], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, 4], [0, 1, 2]], [1, 2, 2, unique-solution, [4/1, 2/1]], [1, 2, 2, unique-solution, [4/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 4], [0, 1, 2]], [1, 2, 2, unique-solution, [4/1, 2/1]], [1, 2, 2, unique-solution, [4/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0093","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7631335aee0675f6b9613f46ac931f73891e9718cd3f770733ef97119753b200","family_id":"qf-density16-system-row-scaling-solution","seed":202610165093,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,5],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 5], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, -10], [0, 1, 2]], [1, 2, 2, unique-solution, [5/1, 2/1]], [-2, 2, 2, unique-solution, [5/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, -10], [0, 1, 2]], [1, 2, 2, unique-solution, [5/1, 2/1]], [-2, 2, 2, unique-solution, [5/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0094","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2462ab353b62dbc93d419a398b8206fffe312d6b0f9305c3a9aa5eadd05f586","family_id":"qf-density16-system-row-scaling-solution","seed":202610165094,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,6],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 6], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, 6], [0, 1, 2]], [1, 2, 2, unique-solution, [6/1, 2/1]], [1, 2, 2, unique-solution, [6/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 6], [0, 1, 2]], [1, 2, 2, unique-solution, [6/1, 2/1]], [1, 2, 2, unique-solution, [6/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0095","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13dfc75a77ea608bad52ae32ea26b0921c11e9a8ae360e654bc016c8da647492","family_id":"qf-density16-system-row-scaling-solution","seed":202610165095,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,7],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 7], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, -14], [0, 1, 2]], [1, 2, 2, unique-solution, [7/1, 2/1]], [-2, 2, 2, unique-solution, [7/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, -14], [0, 1, 2]], [1, 2, 2, unique-solution, [7/1, 2/1]], [-2, 2, 2, unique-solution, [7/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0096","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ef001f30d4e5c301bc8d8a969c42707cf250d2a9ae3f7582a8a7b1a7e20f7c9","family_id":"qf-density16-system-row-scaling-solution","seed":202610165096,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,8],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 8], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, 8], [0, 1, 2]], [1, 2, 2, unique-solution, [8/1, 2/1]], [1, 2, 2, unique-solution, [8/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 8], [0, 1, 2]], [1, 2, 2, unique-solution, [8/1, 2/1]], [1, 2, 2, unique-solution, [8/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0097","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"595e0bc6eb0c4191c7d84e62d53b80870bcf685d5f7359bedd09168932f2b4bd","family_id":"qf-density16-system-row-scaling-solution","seed":202610165097,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,9],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 9], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, -18], [0, 1, 2]], [1, 2, 2, unique-solution, [9/1, 2/1]], [-2, 2, 2, unique-solution, [9/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, -18], [0, 1, 2]], [1, 2, 2, unique-solution, [9/1, 2/1]], [-2, 2, 2, unique-solution, [9/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0098","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ecbc97e478438bb65c5c4eb921963b705c6e8d4b58038dac0d6087590d00286f","family_id":"qf-density16-system-row-scaling-solution","seed":202610165098,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,10],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 10], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, 10], [0, 1, 2]], [1, 2, 2, unique-solution, [10/1, 2/1]], [1, 2, 2, unique-solution, [10/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 10], [0, 1, 2]], [1, 2, 2, unique-solution, [10/1, 2/1]], [1, 2, 2, unique-solution, [10/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0099","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e36f50bdb7a32ccf1be5347899515791d7f70b5cd77616052f0b8d41a3f291b","family_id":"qf-density16-system-row-scaling-solution","seed":202610165099,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,11],[0,1,2]],\"t\":-2}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 11], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=-2; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[-2, 0, -22], [0, 1, 2]], [1, 2, 2, unique-solution, [11/1, 2/1]], [-2, 2, 2, unique-solution, [11/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[-2, 0, -22], [0, 1, 2]], [1, 2, 2, unique-solution, [11/1, 2/1]], [-2, 2, 2, unique-solution, [11/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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-20261012-advance-20261016-r05-system-linear-equations-0100","batch_id":"batch-20261012-advance-20261016-r05-system-linear-equations-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"409059a3a7b3b7a5d15cafee660ad5e4b12dc4c5d9287ee5ecfd2bfcd352be12","family_id":"qf-density16-system-row-scaling-solution","seed":202610165100,"semantic_template":"qf-density16-system-row-scaling-solution/{\"rows\":[[1,0,12],[0,1,2]],\"t\":1}"},"fields":{"title":"Row Scaling: Classification versus Operation Reversibility","stem":"Over R, rows=[[1, 0, 12], [0, 1, 2]], where each [a,b,c] means a*x+b*y=c. t=1; multiply only row1 by t, including its right-side constant. Give (new rows, original classification record [signed D, rankA, rankAb, no-solution/unique-solution/infinitely-many-solutions, ordered rational solution [x,y] or not-applicable/not-unique], new classification record, is t≠0 an invertible operation?, fixed text “Only nonzero scaling is an invertible row operation”). Field 4 is not whether the solution sets happen to be equal.","answer":{"kind":"expression","value":"([[1, 0, 12], [0, 1, 2]], [1, 2, 2, unique-solution, [12/1, 2/1]], [1, 2, 2, unique-solution, [12/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)","explanation":"Recompute both ranks/classes after scaling the whole row; zero scaling can preserve a redundant system’s solution set yet remains a noninvertible row operation. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[1, 0, 12], [0, 1, 2]], [1, 2, 2, unique-solution, [12/1, 2/1]], [1, 2, 2, unique-solution, [12/1, 2/1]], Yes, Only nonzero scaling is an invertible row operation)."},"skills":["Row Scaling: Classification versus Operation Reversibility"],"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."}}