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  "body": "\n## Statement\n\nLet $K$ be a commutative ring and let $A$ be an [[invertible|0EB9N0S33C]] $n \\times n$ matrix over $K$.\nThen $\\det A$ is a unit of $K$, and\n$$\\det(A^{-1}) = (\\det A)^{-1},$$\nwhere $\\det$ is the [[determinant|00KDZRW4C0]].\n\n## Remarks\n\nThe proof applies the [[product rule|04CPJ6AF00]] to $A A^{-1} = I_n$: $\\det A \\cdot \\det(A^{-1}) = 1$.\nOver a field, this says $\\det(A^{-1}) = 1 / \\det A$.\n\n## Explanations\n\n### L1 human\n\nThe inverse matrix undoes the matrix, so it must undo its scaling of volume.\nIf $A$ scales volume by $3$, then $A^{-1}$ scales volume by $1/3$.\nSo the determinant of the inverse is the inverse of the determinant.\n",
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