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  "body": "\n## Statement\n\nLet $K$ be a commutative ring, let $A$ be an $m \\times m$ matrix, $B$ an $m \\times n$ matrix and $D$ an $n \\times n$ matrix over $K$.\nThen the [[block matrix|0ASDFEQS6P]] satisfies\n$$\\det \\begin{pmatrix} A \u0026 B \\\\ 0 \u0026 D \\end{pmatrix} = \\det A \\cdot \\det D,$$\nwhere $0$ is the $n \\times m$ zero matrix and $\\det$ is the [[determinant|00KDZRW4C0]].\n\n## Remarks\n\nThe same holds for the lower block triangular matrix $\\begin{pmatrix} A \u0026 0 \\\\ C \u0026 D \\end{pmatrix}$.\nIt generalizes the rule for [[triangular matrices|0K1KKMFHYH]].\nFor a general block matrix, $\\det \\begin{pmatrix} A \u0026 B \\\\ C \u0026 D \\end{pmatrix} \\ne \\det A \\det D - \\det B \\det C$ in general.\n\n## Explanations\n\n### L1 human\n\nWhen a matrix has a block of zeros in its lower left corner, its determinant splits.\nIt is the product of the determinants of the two diagonal blocks.\nThe block $B$ does not matter, just like the entries above the diagonal of a triangular matrix.\n",
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