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  "body": "\n## Proof\n\nWe compute each entry of the two products by a Laplace expansion.\n\n1. Let $A = (a_{kl})$ be an $n \\times n$ matrix over $K$ with $n \\ge 2$, and let $C_{kl}$ be its [[cofactors|0W2QAVJQ1W]].\n2. By the definition of the [[adjugate|0GMPYM0NSR]], $\\operatorname{adj}(A)_{lk} = C_{kl}$.\n3. By the definition of the [[matrix product|076JGP6R3T]], $(A \\operatorname{adj}(A))_{ik} = \\sum_{j=1}^{n} a_{ij} C_{kj}$ for all $i, k$.\n4. If $i = k$, the sum in step 3 is $\\det A$ by the [[Laplace expansion along row $i$|0C08BQ3M05]].\n5. Let $i \\ne k$. Let $A'$ be the matrix that results when row $k$ of $A$ is replaced by row $i$ of $A$.\n6. The cofactor $C'_{kj}$ of $A'$ deletes row $k$, the only row where $A'$ and $A$ differ. So $C'_{kj} = C_{kj}$ for all $j$.\n7. Row $k$ of $A'$ has entries $a_{ij}$. By the Laplace expansion of $A'$ along row $k$ and step 6, $\\det A' = \\sum_{j=1}^{n} a_{ij} C_{kj}$.\n8. Rows $i$ and $k$ of $A'$ are equal. By the rule for [[two equal rows|0K4VQB7HWR]], $\\det A' = 0$.\n9. By steps 3 to 8, $(A \\operatorname{adj}(A))_{ik}$ is $\\det A$ if $i = k$ and $0$ if $i \\ne k$. So $A \\operatorname{adj}(A) = \\det(A)\\, I_n$, with the [[identity matrix|06DW9BAWB1]] $I_n$.\n10. In the same way, $(\\operatorname{adj}(A) A)_{ik} = \\sum_{j=1}^{n} C_{ji} a_{jk}$ for all $i, k$.\n11. If $i = k$, this sum is $\\det A$ by the [[Laplace expansion along column $i$|06AMCJMMVM]].\n12. Let $i \\ne k$. Let $A''$ be the matrix that results when column $i$ of $A$ is replaced by column $k$ of $A$.\n13. The cofactor $C''_{ji}$ of $A''$ deletes column $i$, the only column where $A''$ and $A$ differ. So $C''_{ji} = C_{ji}$ for all $j$.\n14. Column $i$ of $A''$ has entries $a_{jk}$. By the Laplace expansion of $A''$ along column $i$, $\\det A'' = \\sum_{j=1}^{n} a_{jk} C_{ji}$.\n15. Columns $i$ and $k$ of $A''$ are equal, so the transpose of $A''$ has two equal rows. By the rule for the [[determinant of a transpose|0ZZQ2E1MTE]] and step 8's rule, $\\det A'' = 0$.\n16. By steps 10 to 15, $\\operatorname{adj}(A) A = \\det(A)\\, I_n$. Together with step 9, $A \\operatorname{adj}(A) = \\operatorname{adj}(A) A = \\det(A)\\, I_n$.\n\n## Idea\n\nAn entry of $A \\operatorname{adj}(A)$ pairs one row of $A$ with the cofactors of another row. For the same row this is the Laplace expansion of $\\det A$. For different rows it is the expansion of a matrix with two equal rows, which is $0$.\n",
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