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  "body": "\n## Statement\n\nLet $R$ be a commutative ring and let $A$ and $B$ be [[invertible|0EB9N0S33C]] $n \\times n$ matrices over $R$.\nThen $AB$ is invertible, and\n$$(AB)^{-1} = B^{-1} A^{-1}.$$\n\n## Remarks\n\nThe proof is the computation $(AB)(B^{-1} A^{-1}) = A (B B^{-1}) A^{-1} = A A^{-1} = I_n$, and the same in the other order.\nBy induction, $(A_1 \\cdots A_k)^{-1} = A_k^{-1} \\cdots A_1^{-1}$.\nThus the invertible $n \\times n$ matrices form a group, the [[general linear group|0FCHE902D7]].\n\n## Explanations\n\n### L1 human\n\nTo undo two steps, we undo the last step first.\nIf we first apply $B$ and then $A$, then we must first undo $A$ and then undo $B$.\nSo the inverse of $AB$ is $B^{-1} A^{-1}$, with the order reversed.\n",
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