# Determinant with Rows Transposed

## Theorem

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

## Proof 1

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

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

Let $e$ be the elementary row operation that exchanging rows $r$ and $s$.

Let $\mathbf B = \map e {\mathbf A}$.

Let $\mathbf E$ be the elementary row matrix corresponding to $e$.

$\mathbf B = \mathbf E \mathbf A$
$\map \det {\mathbf E} = -1$

Then:

 $\displaystyle \map \det {\mathbf B}$ $=$ $\displaystyle \map \det {\mathbf E \mathbf A}$ Determinant of Matrix Product $\displaystyle$ $=$ $\displaystyle -\map \det {\mathbf A}$ as $\map \det {\mathbf E} = -1$

Hence the result.

$\blacksquare$

## Proof 2

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 $\rho$ be the permutation on $\N^*_n$ which transposes $r$ and $s$.

From Parity of K-Cycle, $\map \sgn \rho = -1$.

Let $\mathbf A' = \sqbrk {a'}_n$ be $\mathbf A$ with rows $r$ and $s$ transposed.

By the definition of a determinant:

$\displaystyle \map \det {\mathbf A'} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a'_{k \map \lambda k} }$
$\displaystyle \map \det {\mathbf A'} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$

We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:

$\displaystyle \map \det {\mathbf A'} = \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} } = -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$

Hence the result.

$\blacksquare$