Determinant of Elementary Column Matrix/Exchange Columns

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

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 $n$.

By Determinant of Unit Matrix:
 * $\map \det {\mathbf I} = 1$

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

By Permutation of Determinant Indices:
 * $\ds \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }$

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

Hence the result.