Permuting rows multiplies the determinant by the sign

Statement

Let $K$ be a commutative ring, let $A = (a_{ij})$ be an $n \times n$ matrix over $K$, and let $\sigma$ be a permutation of $\{1, \dots, n\}$. Let $B = (b_{ij})$ be the matrix with $b_{ij} = a_{\sigma(i), j}$. Then $\det B = \sgn(\sigma) \det A$, where $\det$ is the determinant.

Remarks

In particular, exchanging two rows changes the sign of the determinant. For a permutation matrix $P$, this gives $\det P = \pm 1$. The same holds for a permutation of the columns.

Explanations

L1 human

When we exchange two rows, the determinant changes its sign. A general reordering of the rows is a sequence of such exchanges. So it multiplies the determinant by $1$ or by $-1$, according to the sign of the permutation.