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  "body": "\n## Statement\n\nLet $R$ be an [[integral domain|0QVD7WBFGN]] and let $A = (a_{ij})$ be an $n \\times n$ matrix over $R$, with $n \\ge 2$.\nPut $a^{(0)}_{ij} = a_{ij}$.\nFor $k = 1, \\dots, n - 1$ and $k \u003c i, j \\le n$, the *Bareiss algorithm* sets\n$$a^{(k)}_{ij} = \\frac{a^{(k-1)}_{kk} \\, a^{(k-1)}_{ij} - a^{(k-1)}_{ik} \\, a^{(k-1)}_{kj}}{a^{(k-2)}_{k-1, k-1}},$$\nwhere the denominator is $1$ for $k = 1$.\nAssume that the leading principal [[minors|0BT9BYWVRY]] of $A$ of orders $1, \\dots, n - 1$ are nonzero.\nThen:\n\n1. for each $k$ and each $i, j \u003e k$, the entry $a^{(k)}_{ij}$ equals the determinant of the $(k+1) \\times (k+1)$ submatrix of $A$ with rows $1, \\dots, k, i$ and columns $1, \\dots, k, j$;\n2. in particular, every division in the algorithm is exact in $R$, every pivot $a^{(k-1)}_{kk}$ is the leading principal minor of order $k$, and $a^{(n-1)}_{nn} = \\det A$.\n\n## Remarks\n\nClause 1 is proved by induction on $k$ with Sylvester's determinant identity, which expresses a bordered minor of the reduced matrix as a ratio of minors of $A$.\nFor $k = 1$ it is the direct check $a^{(1)}_{ij} = a_{11} a_{ij} - a_{i1} a_{1j} = \\det \\begin{pmatrix} a_{11} \u0026 a_{1j} \\\\ a_{i1} \u0026 a_{ij} \\end{pmatrix}$.\nExample: for $A = \\begin{pmatrix} 2 \u0026 3 \u0026 1 \\\\ 4 \u0026 1 \u0026 5 \\\\ 6 \u0026 2 \u0026 3 \\end{pmatrix}$ over $\\Z$, step $1$ gives $a^{(1)}_{22} = -10$, $a^{(1)}_{23} = 6$, $a^{(1)}_{32} = -14$, $a^{(1)}_{33} = 0$, and step $2$ gives $a^{(2)}_{33} = \\big((-10) \\cdot 0 - 6 \\cdot (-14)\\big)/2 = 42 = \\det A$.\nOver $\\Z$, clause 1 and [[Hadamard's inequality|CGDN1X32NQ]] bound the size of every intermediate entry.\nSo the bit length of the numbers grows only polynomially, unlike naive fraction-free elimination, where it can double at each step.\nIf a pivot is zero, rows can be exchanged as in [[Gaussian elimination|0JVJ2WCXZH]], which changes the sign of the [[determinant|00KDZRW4C0]].\nThe algorithm uses $O(n^3)$ operations in $R$.\n\n## Explanations\n\n### L1 human\n\nTo compute a determinant of an integer matrix exactly, one wants to avoid fractions.\nCross-multiplying rows avoids fractions but makes the numbers huge.\nThe Bareiss algorithm cross-multiplies and then divides by the previous pivot, and a theorem guarantees that this division always leaves no remainder.\n\n### L2 human\n\nIt is fraction-free Gaussian elimination: the work is the same order as ordinary elimination, but all intermediate values are minors of the input matrix.\nIt is a standard method in computer algebra for determinants, ranks and exact solutions of linear systems over $\\Z$ or over polynomial rings.\nThe final pivot is the determinant, and the earlier pivots are the leading principal minors.\n",
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