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  "body": "\n## Statement\n\nLet $F$ be a field and let $A$ and $B$ be $n \\times n$ matrices over $F$.\nIf $BA = I_n$, then $AB = I_n$, so $A$ is [[invertible|0EB9N0S33C]] and $A^{-1} = B$.\nThe same holds with the roles of $A$ and $B$ exchanged.\n\n## Remarks\n\nThe statement also holds over each commutative ring, by the [[determinant|00KDZRW4C0]]: $BA = I_n$ gives $\\det B \\det A = 1$.\nFor non-square matrices it fails: a $2 \\times 1$ matrix can have a left inverse but never a right inverse.\nA proof over a field uses [[rank–nullity|0W4FSW1TVB]]: $BA = I_n$ makes $x \\mapsto Ax$ injective, hence surjective.\n\n## Explanations\n\n### L1 human\n\nFor square matrices, we only need to check one of $AB = I$ and $BA = I$.\nThe other one then holds automatically.\nThis saves half of the work when we check an inverse.\n",
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