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  "body": "\n## Statement\n\nLet $K$ be a [[commutative ring|0310REJJY8]], let $1 \\le k \u003c n$, and let $A = (a_{ij})$ be an $n \\times n$ matrix over $K$.\nLet $A_0$ be the $k \\times k$ submatrix of $A$ formed by its first $k$ rows and its first $k$ columns.\nFor $k \u003c i, j \\le n$, let $b_{ij}$ be the [[minor|0BT9BYWVRY]] of $A$ with rows $1, \\dots, k, i$ and columns $1, \\dots, k, j$.\nLet $B$ be the $(n-k) \\times (n-k)$ matrix $(b_{ij})_{k \u003c i, j \\le n}$.\nThen\n$$\\det B = (\\det A_0)^{n-k-1} \\det A,$$\nwhere $\\det$ is the [[determinant|00KDZRW4C0]].\n\n## Remarks\n\nFirst assume that $A_0$ is invertible, and write $A$ in blocks with $A_0$ in the top left corner.\nBy the [[Schur complement formula|0KA7A4ENVJ]], $\\det A = \\det A_0 \\cdot \\det S$, where $S = (s_{ij})$ is the Schur complement of $A_0$ in $A$.\nThe same formula on each bordered submatrix gives $b_{ij} = \\det A_0 \\cdot s_{ij}$, so $B = (\\det A_0) S$ and the identity follows.\nBoth sides are polynomials in the entries of $A$ with integer coefficients, so the identity holds over every commutative ring.\n\nFor $k = n - 1$, the matrix $B$ is $1 \\times 1$ with entry $\\det A$, and the identity is trivial.\nFor $k = n - 2$, it is a form of the Desnanot–Jacobi identity, which is the basis of [[Dodgson condensation|RQ44FKS5QM]].\nIt should not be confused with [[Sylvester's determinant theorem|0G24VQ1088]], which says $\\det(I_m + AB) = \\det(I_n + BA)$.\n\n## Explanations\n\n### L1 human\n\nFix the top left corner block of a square matrix.\nBuild a smaller matrix whose entries are determinants of this corner block bordered by one more row and one more column.\nThe determinant of the new matrix equals the determinant of the original matrix times a power of the determinant of the corner block.\nFor a $3 \\times 3$ matrix and $k = 1$, the four entries of $B$ are $2 \\times 2$ minors that all contain the entry $a_{11}$, and $\\det B = a_{11} \\det A$.\n\n### L2 human\n\nThe identity explains why fraction-free Gaussian elimination works over the integers.\nThe intermediate entries of the Bareiss algorithm are such bordered minors, and the division by the previous pivot is exact.\nIt sits next to the [[determinant via the Schur complement|0KA7A4ENVJ]] and to the [[determinant of a block triangular matrix|0J9TVTNB4K]].\n",
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