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  "body": "\n## Proof\n\nWe group the terms of the Leibniz formula by the column of the factor taken from row $i$.\n\n1. Let $A = (a_{kl})$ be an $n \\times n$ matrix over $K$ with $n \\ge 2$, and fix a row $i$.\n2. By the [[Leibniz formula|016PE4VZ8Q]], $\\det A = \\sum_{\\tau \\in S_n} \\sgn(\\tau) \\prod_{k=1}^{n} a_{k, \\tau(k)}$, where $S_n$ is the set of [[permutations|05RWFKVTJ7]] of $\\set{1, \\dots, n}$.\n3. Each $\\tau \\in S_n$ has exactly one value $\\tau(i)$. Sort the terms by $j = \\tau(i)$ and take out the factor $a_{ij}$.\n4. This gives $\\det A = \\sum_{j=1}^{n} a_{ij} T_j$, where $T_j = \\sum_{\\tau \\in S_n,\\ \\tau(i) = j} \\sgn(\\tau) \\prod_{k \\ne i} a_{k, \\tau(k)}$.\n5. By the definition of the [[cofactor|0W2QAVJQ1W]], it is enough to show $T_j = (-1)^{i+j} M_{ij}$ for each $j$. Fix $j$.\n6. Let $\\alpha \\colon \\set{1, \\dots, n-1} \\to \\set{1, \\dots, n} \\setminus \\set{i}$ be the increasing bijection. So $\\alpha(r) = r$ for $r \u003c i$ and $\\alpha(r) = r + 1$ for $r \\ge i$.\n7. Let $\\beta \\colon \\set{1, \\dots, n-1} \\to \\set{1, \\dots, n} \\setminus \\set{j}$ be the increasing bijection in the same way.\n8. Let $A_{ij}$ be $A$ with row $i$ and column $j$ deleted. Its entry in position $(r, s)$ is $a_{\\alpha(r), \\beta(s)}$. By the definition of the [[minor|0SD528P0EK]], $M_{ij} = \\det A_{ij}$.\n9. Let $\\tau \\in S_n$ with $\\tau(i) = j$. Then $\\tau$ maps $\\set{1, \\dots, n} \\setminus \\set{i}$ bijectively onto $\\set{1, \\dots, n} \\setminus \\set{j}$.\n10. So $\\sigma = \\beta^{-1} \\circ \\tau \\circ \\alpha$ is a permutation of $\\set{1, \\dots, n-1}$.\n11. Conversely, each $\\sigma \\in S_{n-1}$ comes from exactly one such $\\tau$. It is given by $\\tau(i) = j$ and $\\tau(\\alpha(r)) = \\beta(\\sigma(r))$ for $1 \\le r \\le n-1$.\n12. So $\\tau \\mapsto \\sigma$ is a bijection from $\\set{\\tau \\in S_n : \\tau(i) = j}$ onto $S_{n-1}$.\n13. For such $\\tau$ and $\\sigma$, the indices $k \\ne i$ are exactly the values $\\alpha(r)$. So $\\prod_{k \\ne i} a_{k, \\tau(k)} = \\prod_{r=1}^{n-1} a_{\\alpha(r), \\beta(\\sigma(r))}$.\n14. By step 8, this is $\\prod_{r=1}^{n-1} (A_{ij})_{r, \\sigma(r)}$.\n15. Now we compare signs. Extend $\\alpha$ to $\\hat\\alpha \\in S_n$ by $\\hat\\alpha(n) = i$. It sends $i \\mapsto i+1 \\mapsto \\cdots \\mapsto n \\mapsto i$ and fixes the other points.\n16. So $\\hat\\alpha$ is a cycle of length $n - i + 1$. By the [[sign of a cycle|0V0E04DWX2]], $\\sgn(\\hat\\alpha) = (-1)^{n-i}$.\n17. In the same way, extend $\\beta$ to $\\hat\\beta \\in S_n$ by $\\hat\\beta(n) = j$. Then $\\sgn(\\hat\\beta) = (-1)^{n-j}$.\n18. Let $\\hat\\sigma = \\hat\\beta^{-1} \\circ \\tau \\circ \\hat\\alpha$. Then $\\hat\\sigma(n) = \\hat\\beta^{-1}(\\tau(i)) = \\hat\\beta^{-1}(j) = n$.\n19. For $r \\le n-1$, $\\tau(\\alpha(r)) \\ne j$, so $\\hat\\sigma(r) = \\beta^{-1}(\\tau(\\alpha(r))) = \\sigma(r)$. So $\\hat\\sigma$ fixes $n$ and agrees with $\\sigma$ elsewhere.\n20. Write $\\sigma$ as a product of $m$ transpositions of $\\set{1, \\dots, n-1}$, with $m = 0$ if $\\sigma$ is the identity. The same product, read in $S_n$, is $\\hat\\sigma$.\n21. By the definition of the [[sign|0SK90238BH]], $\\sgn(\\hat\\sigma) = (-1)^m = \\sgn(\\sigma)$.\n22. Since $\\tau = \\hat\\beta \\circ \\hat\\sigma \\circ \\hat\\alpha^{-1}$ and the [[sign is a homomorphism|0JEJJCB0EX]], $\\sgn(\\tau) = \\sgn(\\hat\\beta) \\sgn(\\sigma) \\sgn(\\hat\\alpha)^{-1}$.\n23. By steps 16 and 17, $\\sgn(\\tau) = (-1)^{n-j} (-1)^{n-i} \\sgn(\\sigma) = (-1)^{i+j} \\sgn(\\sigma)$.\n24. By steps 12, 14 and 23, $T_j = (-1)^{i+j} \\sum_{\\sigma \\in S_{n-1}} \\sgn(\\sigma) \\prod_{r=1}^{n-1} (A_{ij})_{r, \\sigma(r)}$.\n25. By the Leibniz formula for the $(n-1) \\times (n-1)$ matrix $A_{ij}$, the sum in step 24 is $\\det A_{ij} = M_{ij}$. So $T_j = (-1)^{i+j} M_{ij} = C_{ij}$.\n26. By step 4, $\\det A = \\sum_{j=1}^{n} a_{ij} C_{ij} = \\sum_{j=1}^{n} (-1)^{i+j} a_{ij} M_{ij}$.\n\n## Idea\n\nEvery term of the Leibniz formula uses exactly one entry of row $i$. The terms that use $a_{ij}$ are, after renumbering, the terms of the Leibniz formula for the minor $M_{ij}$. The renumbering moves row $i$ and column $j$ to the end, and this costs the sign $(-1)^{i+j}$.\n",
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