Proof that permuting rows multiplies the determinant by the sign

Proof

We first treat one exchange of two rows, and then write $\sigma$ as a product of transpositions.

1. 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 $B = A_\sigma$.
2. 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 $(M_\pi)_\rho = M_{\pi \rho}$.
3. Let $\tau$ be the transposition that exchanges $k$ and $l$, with $k \ne l$. Then $M_\tau$ is $M$ with rows $k$ and $l$ exchanged. We show $\det M_\tau = -\det M$.
4. 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 $v$.
5. Then $D(x, y) = \det M$ and $D(y, x) = \det M_\tau$. By the rule for two equal rows, $D(u, u) = 0$ for every row $u$.
6. By multilinearity in row $k$ and then in row $l$, $D(x + y, x + y) = D(x, x) + D(x, y) + D(y, x) + D(y, y)$.
7. 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, $\det M_\tau = -\det M$.
8. If $n = 1$, then $\sigma$ is the identity, $B = A$ and $\sgn(\sigma) = 1$. So the claim holds. From now on let $n \ge 2$.
9. The symmetric group $S_n$ is generated by transpositions, and the inverse of a transposition is itself. So $\sigma = \tau_1 \tau_2 \cdots \tau_m$ with transpositions $\tau_r$.
10. 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 $A_{\pi_m} = B$.
11. By step 2, $A_{\pi_r} = (A_{\pi_{r-1}})_{\tau_r}$ for $1 \le r \le m$. So step 7 gives $\det A_{\pi_r} = -\det A_{\pi_{r-1}}$.
12. By induction on $r$, $\det A_{\pi_r} = (-1)^r \det A$. For $r = m$, this is $\det B = (-1)^m \det A$.
13. By the definition of the sign, $(-1)^m = \sgn(\sigma)$. So $\det B = \sgn(\sigma) \det A$.