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  "body": "\n## Proof\n\nWe first treat one exchange of two rows, and then write $\\sigma$ as a product of transpositions.\n\n1. For a matrix $M$ and a permutation $\\pi$ of $\\set{1, \\dots, n}$, let $M_\\pi$ be the matrix whose row $i$ is row $\\pi(i)$ of $M$. Then $B = A_\\sigma$.\n2. For permutations $\\pi$ and $\\rho$, row $i$ of $(M_\\pi)_\\rho$ is row $\\rho(i)$ of $M_\\pi$. This is row $\\pi(\\rho(i))$ of $M$. So $(M_\\pi)_\\rho = M_{\\pi \\rho}$.\n3. Let $\\tau$ be the [[transposition|0FWVRHBRWM]] that exchanges $k$ and $l$, with $k \\ne l$. Then $M_\\tau$ is $M$ with rows $k$ and $l$ exchanged. We show $\\det M_\\tau = -\\det M$.\n4. Let $x$ and $y$ be rows $k$ and $l$ of $M$. For rows $u, v$, let $D(u, v)$ be the determinant of $M$ with row $k$ replaced by $u$ and row $l$ replaced by $v$.\n5. Then $D(x, y) = \\det M$ and $D(y, x) = \\det M_\\tau$. By the rule for [[two equal rows|0K4VQB7HWR]], $D(u, u) = 0$ for every row $u$.\n6. By [[multilinearity|0PPG793642]] in row $k$ and then in row $l$, $D(x + y, x + y) = D(x, x) + D(x, y) + D(y, x) + D(y, y)$.\n7. By step 5, the left side and the terms $D(x, x)$ and $D(y, y)$ of step 6 are $0$. So $D(y, x) = -D(x, y)$, that is, $\\det M_\\tau = -\\det M$.\n8. If $n = 1$, then $\\sigma$ is the identity, $B = A$ and $\\sgn(\\sigma) = 1$. So the claim holds. From now on let $n \\ge 2$.\n9. The symmetric group $S_n$ is [[generated by transpositions|0AQNX7G443]], and the inverse of a transposition is itself. So $\\sigma = \\tau_1 \\tau_2 \\cdots \\tau_m$ with transpositions $\\tau_r$.\n10. Let $\\pi_0$ be the identity and $\\pi_r = \\tau_1 \\cdots \\tau_r$ for $1 \\le r \\le m$. Then $A_{\\pi_0} = A$ and $A_{\\pi_m} = B$.\n11. By step 2, $A_{\\pi_r} = (A_{\\pi_{r-1}})_{\\tau_r}$ for $1 \\le r \\le m$. So step 7 gives $\\det A_{\\pi_r} = -\\det A_{\\pi_{r-1}}$.\n12. By induction on $r$, $\\det A_{\\pi_r} = (-1)^r \\det A$. For $r = m$, this is $\\det B = (-1)^m \\det A$.\n13. By the [[definition of the sign|0SK90238BH]], $(-1)^m = \\sgn(\\sigma)$. So $\\det B = \\sgn(\\sigma) \\det A$.\n\n## Idea\n\nPutting the sum of two rows into both places gives a determinant zero. Expanding it shows that one exchange of two rows changes the sign. A general permutation of the rows is a chain of exchanges, so the determinant picks up one factor $-1$ for each.\n",
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