# Determinant with Columns Transposed

## Theorem

If two columns of a matrix with determinant $D$ are transposed, its determinant becomes $-D$.

## Proof

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Let $1 \le r < s \le n$.

Let $\mathbf B$ be $\mathbf A$ with columns $r$ and $s$ transposed.

Consider:

the transpose $\mathbf A^\intercal$ of $\mathbf A$
the transpose $\mathbf B^\intercal$ of $\mathbf B$.

hence $\mathbf B^\intercal$ is $\mathbf A^\intercal$ with rows $r$ and $s$ transposed.

$\map \det {\mathbf B^\intercal} = -\map \det {\mathbf A^\intercal}$

From from Determinant of Transpose:

$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$
$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$

and the result follows.

$\blacksquare$