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  "body": "\n## Statement\n\nLet $K$ be a commutative ring and let $A$ be an invertible $p \\times p$ matrix over $K$.\nLet $B$, $C$ and $D$ be matrices over $K$ of sizes $p \\times q$, $q \\times p$ and $q \\times q$.\nThen\n$$\\det \\begin{pmatrix} A \u0026 B \\\\ C \u0026 D \\end{pmatrix} = \\det A \\cdot \\det(D - C A^{-1} B),$$\nwhere $\\det$ is the [[determinant|00KDZRW4C0]].\nThe matrix $D - C A^{-1} B$ is the *Schur complement* of $A$ in the [[block matrix|0ASDFEQS6P]].\n\n## Remarks\n\nThe proof factors the block matrix as $\\begin{pmatrix} I \u0026 0 \\\\ C A^{-1} \u0026 I \\end{pmatrix} \\begin{pmatrix} A \u0026 B \\\\ 0 \u0026 D - C A^{-1} B \\end{pmatrix}$.\nThen it uses the rule for [[block triangular matrices|0J9TVTNB4K]].\nFor $p = q = 1$ and $a \\ne 0$, it gives $\\det \\begin{pmatrix} a \u0026 b \\\\ c \u0026 d \\end{pmatrix} = a(d - cb/a) = ad - bc$.\n\n## Explanations\n\n### L1 human\n\nBlock elimination removes the block $C$, in the same way as Gaussian elimination removes an entry.\nWhat remains in the lower right corner is the Schur complement.\nThe determinant is then the product of two smaller determinants.\n",
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