{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-matrix-rank-01",
  "mko_id": "mko-matrix-rank",
  "title": {
    "zh-Hant": "三階矩陣的完整RREF與上下界",
    "en": "Full RREF and rank bounds"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-matrix-rank-01",
    "source": {
      "mko_id": "mko-matrix-rank",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "fdddbb93e9e8a00070f3e18fbc7fb5a9406d8ddab249ec9f7798986636ffd45e",
      "english_sha256": "a23c09f0694655deaf0ccb46613f70ba13f76194f5d465dd35153804fabb5727"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "A=[[1,2,3],[2,4,6],[1,1,1]]的row與column rank均為2。",
      "en": "A=[[1,2,3],[2,4,6],[1,1,1]] has row and column rank 2."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "A三rows為(1,2,3)、(2,4,6)、(1,1,1)，在R上計rank。",
          "en": "A has rows (1,2,3),(2,4,6),(1,1,1), with rank over R."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "A=\\begin{pmatrix}1&2&3\\\\2&4&6\\\\1&1&1\\end{pmatrix}",
        "after_tex": "R_2\\leftarrow R_2-2R_1:\\quad\\begin{pmatrix}1&2&3\\\\0&0&0\\\\1&1&1\\end{pmatrix}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "可逆消去清掉第二row；加2R₁可回復。",
          "en": "Invertible elimination clears row two; adding 2R1 reverses it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "零row不表示整個矩陣為零或rank0。",
            "en": "A zero row does not make the matrix zero or rank zero."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "\\begin{pmatrix}1&2&3\\\\0&0&0\\\\1&1&1\\end{pmatrix}",
        "after_tex": "R_3\\leftarrow R_3-R_1:\\quad\\begin{pmatrix}1&2&3\\\\0&0&0\\\\0&-1&-2\\end{pmatrix}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第三row減第一row，得到第二個非零pivot位置。",
          "en": "Subtract row one from row three to expose the second nonzero pivot."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "運算不改row span，新舊rows互為線性組合。",
            "en": "Row span is unchanged because old and new rows generate each other."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "\\begin{pmatrix}1&2&3\\\\0&0&0\\\\0&-1&-2\\end{pmatrix}",
        "after_tex": "R_2\\leftrightarrow R_3,\\quad R_2\\leftarrow-R_2:\\quad\\begin{pmatrix}1&2&3\\\\0&1&2\\\\0&0&0\\end{pmatrix}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交換rows與乘非零−1皆可逆，把pivot正規化。",
          "en": "Row swapping and scaling by nonzero -1 are invertible and normalize the pivot."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "未除以零；−1的逆數仍為−1。",
            "en": "No zero division occurs; the inverse of -1 is -1."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "\\begin{pmatrix}1&2&3\\\\0&1&2\\\\0&0&0\\end{pmatrix}",
        "after_tex": "R_1\\leftarrow R_1-2R_2:\\quad\\operatorname{RREF}(A)=\\begin{pmatrix}1&0&-1\\\\0&1&2\\\\0&0&0\\end{pmatrix}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "清掉第二pivot上方的2，全部rows已至RREF。",
          "en": "Clearing the 2 above pivot two gives the full RREF."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "兩非零rows有不同pivot，獨立並生成全部row space。",
            "en": "The two nonzero rows have distinct pivots, are independent and span the row space."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01",
          "S02",
          "S03",
          "S04"
        ],
        "before_tex": "r_1=(1,0,-1),\\quad r_2=(0,1,2)",
        "after_tex": "\\alpha r_1+\\beta r_2=0\\Rightarrow\\alpha=\\beta=0;\\quad\\operatorname{rowrank}A=2",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "前兩座標給係數零；可逆row操作保留row span，故原矩陣row rank2。",
          "en": "The first two coordinates force zero coefficients; invertible operations preserve row span, giving rank 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S02",
            "S03",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "完整生成與獨立同時給上下界，不是只數非零rows。",
            "en": "Full spanning and independence give both bounds, not just a count of nonzero rows."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1"
        ],
        "before_tex": "c_1=(1,2,1),\\quad c_2=(2,4,1),\\quad c_3=(3,6,1)",
        "after_tex": "c_3=-c_1+2c_2\\Rightarrow\\operatorname{colrank}A\\le2",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第三column由前兩生成，完整描述全部columns，給上界2。",
          "en": "The third column is generated by the first two, covering all columns and giving upper bound 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "非零minor只給下界，此關係才給上界。",
            "en": "A nonzero minor gives only a lower bound; this relation supplies the upper bound."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S06"
        ],
        "before_tex": "\\begin{pmatrix}1&2\\\\1&1\\end{pmatrix}\\text{ in rows 1,3 and columns 1,2}",
        "after_tex": "\\det=1-2=-1\\ne0\\Rightarrow c_1,c_2\\text{ independent}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "關係αc₁+βc₂=0限制到第一、三row給可逆2×2系統，故係數零。",
          "en": "A relation on c1,c2 restricts to an invertible system in rows one and three, forcing zero coefficients."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "minor只認證兩columns獨立；需S06才確定rank。",
            "en": "The minor certifies independence only; S06 is needed for exact rank."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S05",
          "S06",
          "S07"
        ],
        "before_tex": "2\\le\\operatorname{colrank}A\\le2,\\quad\\operatorname{rowrank}A=2",
        "after_tex": "\\operatorname{rank}A=2,\\quad\\det A=0",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "直接證此例兩rank同為2；第二row與第一成比例故det0，但det0單獨不給rank2。",
          "en": "Both ranks are directly 2 here. Proportional rows give determinant zero, but determinant zero alone does not determine rank 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S05",
            "S06",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "未用此例證一般row-rank等於column-rank定理。",
            "en": "This example does not prove the general row/column-rank theorem."
          }
        },
        "leaf_basis": "mko_definition"
      }
    ],
    "result": {
      "zh-Hant": "A=[[1,2,3],[2,4,6],[1,1,1]]的row與column rank均為2。",
      "en": "A=[[1,2,3],[2,4,6],[1,1,1]] has row and column rank 2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "det A=0故rank A=0。",
          "en": "det A=0 implies rank A=0."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "非零2×2 minor給rank至少2。",
          "en": "A nonzero 2-by-2 minor gives rank at least 2."
        }
      },
      {
        "claim": {
          "zh-Hant": "非零2×2 minor已給rank上界2。",
          "en": "A nonzero 2-by-2 minor gives upper bound 2."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "I₃也有這種minor但rank3。",
          "en": "I3 has such a minor but rank 3."
        }
      },
      {
        "claim": {
          "zh-Hant": "三columns皆非零所以rank3。",
          "en": "Three nonzero columns imply rank 3."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "c₃=−c₁+2c₂為非平凡關係。",
          "en": "c3=-c1+2c2 is a nontrivial relation."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設通常實數域與座標向量運算、矩陣乘法、線性獨立及生成的定義；所用消去均為可逆初等列運算。",
          "en": "Assume the real field, usual coordinate-vector arithmetic, matrix multiplication, and definitions of independence and span; elimination uses invertible elementary row operations."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "本輪處理2×2或3×3整數矩陣。非零r-minor只給rank下界；rank上界須全部較大minor為零或完整exact RREF。det=0只排除full rank，不單獨決定rank；row-rank=column-rank依定理。",
          "en": "This slice handles 2-by-2 or 3-by-3 integer matrices. A nonzero r-minor gives only a lower bound. An upper bound requires all larger minors to vanish or a complete exact RREF. A zero determinant excludes full rank without determining the exact lower rank; equality of row rank and column rank is a theorem."
        },
        "source": {
          "mko_id": "mko-matrix-rank",
          "sha256": "fdddbb93e9e8a00070f3e18fbc7fb5a9406d8ddab249ec9f7798986636ffd45e",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "minor",
        "lhs": "1*1-2*1",
        "rhs": "-1",
        "step": "S07"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "完整3×3整數矩陣，精確消去及直接column-span證書。",
      "en": "A complete integer 3-by-3 matrix, exact elimination and a direct column-span certificate."
    },
    "md_symbols": [
      [
        "r_i",
        "RREF非零rows",
        "nonzero RREF rows"
      ],
      [
        "c_i",
        "原columns",
        "original columns"
      ]
    ],
    "md_error_controls": [
      {
        "title": "可逆操作 / Invertible operations",
        "zh": "每步可逆，精確保持row span。",
        "en": "Every step is invertible, preserving the exact row span."
      },
      {
        "title": "雙界 / Both bounds",
        "zh": "非零 minor 給下界；完整的 column-span 證書或完整消去給上界。",
        "en": "A nonzero minor gives a lower bound; a full column-span certificate or complete elimination gives an upper bound."
      }
    ],
    "md_open_items": []
  },
  "revision": {
    "id": "ocme-new142-reader-adoption-20261021",
    "collection": "coverage-20261021",
    "source_commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
    "adopted_source_sha256": "d8af8464c4da76d69787d86c35189042d83bb8445d6d889fd803182b71cf72bb",
    "original_source_sha256": "68e30ccbe7cb5fcb0a3a995a4a766df4bcf2e97e70abc01d7195859973bf2fe5",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
    "independent_AI_concurrence": "SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW",
    "issue_ids": [
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    ],
    "parent_manifest_sha256": "d29f4cae8229ebd24f60d496cb1d3cd4ad8c0a4dca9d44daab0c3245c544283a",
    "scope": "Owner-adopted reader revision. Lead and bilingual review concern the recorded text and premises; no new universal or Lean proof, canonical promotion or historical-label rewrite."
  }
}