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  "body": "\n## Proof\n\nWe divide the identity for the adjugate by the unit $\\det A$.\n\n1. Let $A$ be an $n \\times n$ matrix over $K$ with $n \\ge 2$. Let $d = \\det A$, and assume that $d$ is a [[unit|019NFGFPR5]] of $K$.\n2. Let $B = d^{-1} \\operatorname{adj}(A)$, where $\\operatorname{adj}(A)$ is the [[adjugate|0GMPYM0NSR]] of $A$.\n3. By the definition of the [[matrix product|076JGP6R3T]] and the commutativity of $K$, a scalar factor can be moved out of a product: $A (c M) = c (A M)$ and $(c M) A = c (M A)$.\n4. So $A B = d^{-1} (A \\operatorname{adj}(A))$ and $B A = d^{-1} (\\operatorname{adj}(A) A)$.\n5. By the [[product of a matrix and its adjugate|0528HKWZ1Z]], $A \\operatorname{adj}(A) = \\operatorname{adj}(A) A = d I_n$.\n6. By steps 4 and 5, $A B = d^{-1} d I_n = I_n$ and $B A = I_n$.\n7. By the definition of an [[invertible matrix|0TWTS88TWN]], $A$ is invertible, and its inverse is unique. So $A^{-1} = B = (\\det A)^{-1} \\operatorname{adj}(A)$.\n\n## Idea\n\nThe adjugate already inverts $A$ up to the scalar factor $\\det A$. When this scalar has an inverse in $K$, dividing by it gives the inverse matrix on both sides.\n",
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