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  "body": "\n## Proof\n\nWe apply the expansion along a row to the transpose of $A$.\n\n1. Let $A = (a_{kl})$ be an $n \\times n$ matrix over $K$ with $n \\ge 2$, and fix a column $j$.\n2. Let $B = A^T$ be the [[transpose|0W809K6AW0]] of $A$. So $B$ has entries $b_{kl} = a_{lk}$.\n3. For indices $r, s$, let $N_{rs}$ be the [[minor|0SD528P0EK]] of $B$ that results when row $r$ and column $s$ of $B$ are deleted.\n4. Fix $i$, and let $A_{ij}$ be $A$ with row $i$ and column $j$ deleted. Deleting row $j$ and column $i$ of $B$ gives the transpose $(A_{ij})^T$.\n5. Indeed, both matrices keep the entries $a_{kl}$ with $k \\ne i$ and $l \\ne j$, in the same order. Each puts $a_{kl}$ in the row for $l$ and the column for $k$.\n6. By the rule for the [[determinant of a transpose|0ZZQ2E1MTE]], $N_{ji} = \\det (A_{ij})^T = \\det A_{ij} = M_{ij}$, for each $i$.\n7. By the [[Laplace expansion along row $j$|0C08BQ3M05]] of $B$, $\\det B = \\sum_{i=1}^{n} (-1)^{j+i} b_{ji} N_{ji}$.\n8. By steps 2 and 6, the right side is $\\sum_{i=1}^{n} (-1)^{i+j} a_{ij} M_{ij}$.\n9. By the rule for the determinant of a transpose, $\\det B = \\det A$.\n10. With the [[cofactors|0W2QAVJQ1W]] $C_{ij} = (-1)^{i+j} M_{ij}$, steps 7 to 9 give $\\det A = \\sum_{i=1}^{n} a_{ij} C_{ij} = \\sum_{i=1}^{n} (-1)^{i+j} a_{ij} M_{ij}$.\n\n## Idea\n\nTransposing turns columns into rows and does not change any determinant. So the expansion along column $j$ of $A$ is the expansion along row $j$ of $A^T$, with the same minors.\n",
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