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  "body": "\n## Statement\n\nLet $K$ be a commutative ring, let $n \\ge 2$, and let $A$ be an $n \\times n$ matrix over $K$ such that $\\det A$ is a unit of $K$.\nThen $A$ is [[invertible|0EB9N0S33C]], and\n$$A^{-1} = (\\det A)^{-1} \\operatorname{adj}(A),$$\nwhere $\\operatorname{adj}(A)$ is the [[adjugate|07W9H8QF38]] of $A$.\n\n## Remarks\n\nThe statement follows by dividing [[$A \\operatorname{adj}(A) = \\operatorname{adj}(A) A = \\det(A) I_n$|0528HKWZ1Z]] by $\\det A$.\nFor $n = 2$ and $ad - bc \\ne 0$ over a field, $\\begin{pmatrix} a \u0026 b \\\\ c \u0026 d \\end{pmatrix}^{-1} = \\frac{1}{ad - bc} \\begin{pmatrix} d \u0026 -b \\\\ -c \u0026 a \\end{pmatrix}$.\nFor large $n$, row reduction computes the inverse much faster than this formula.\n\n## Explanations\n\n### L1 human\n\nThis formula gives every entry of the inverse as a quotient of two determinants.\nIt is useful for theory and for small matrices.\nFor $2 \\times 2$ matrices, it is the formula that most students learn first.\n",
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