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  "body": "\n## Statement\n\nLet $K$ be a commutative ring, let $n \\ge 1$, and let $A = (a_{ij})$ be an $n \\times n$ matrix over $K$.\nLet $S_n$ be the group of permutations of $\\{1, \\dots, n\\}$, and let $\\sgn(\\sigma) \\in \\{1, -1\\}$ be the sign of $\\sigma$.\nThen the [[determinant|00KDZRW4C0]] of $A$ is\n$$\\det A = \\sum_{\\sigma \\in S_n} \\sgn(\\sigma) \\prod_{i=1}^{n} a_{i, \\sigma(i)}.$$\n\n## Remarks\n\nThe sum has $n!$ terms.\nEach term takes exactly one entry from each row and from each column.\nFor $n = 3$, the formula is the [[rule of Sarrus|0H8SZFFDXJ]].\nThe formula shows that the determinant is a polynomial in the entries, with integer coefficients.\n\n## Explanations\n\n### L1 human\n\nTo compute a determinant by this formula, choose one entry in each row so that no two lie in the same column.\nMultiply the chosen entries, and give the product the sign of the permutation.\nAdd these signed products over all choices.\n",
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