# Determinant of Elementary Column Matrix/Exchange Columns

## Theorem

Let $e_3$ be the elementary column operation $\text {ECO} 3$:

 $(\text {ECO} 3)$ $:$ $\displaystyle \kappa_i \leftrightarrow \kappa_j$ Exchange columns $i$ and $j$

which is to operate on some arbitrary matrix space.

Let $\mathbf E_3$ be the elementary column matrix corresponding to $e_3$.

The determinant of $\mathbf E_3$ is:

$\map \det {\mathbf E_3} = -1$

## Proof

Let $\mathbf I$ denote the unit matrix of arbitrary order $n$.

$\map \det {\mathbf I} = 1$

Let $\rho$ be the permutation on $\tuple {1, 2, \ldots, n}$ which transposes $i$ and $j$.

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

By definition we have that $\mathbf E_3$ is $\mathbf I$ with columns $i$ and $j$ transposed.

By the definition of a determinant:

$\displaystyle \map \det {\mathbf I} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }$
$\displaystyle \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }$

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

 $\ds \map \det {\mathbf E_3}$ $=$ $\ds \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }$ $\ds$ $=$ $\ds -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }$ $\ds$ $=$ $\ds -\map \det {\mathbf I}$

Hence the result.

$\blacksquare$