# Determinant of Elementary Row Matrix/Exchange Rows

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## Theorem

Let $e_3$ be the elementary row operation $\text {ERO} 3$:

 $(\text {ERO} 3)$ $:$ $\ds r_i \leftrightarrow r_j$ Exchange rows $i$ and $j$

which is to operate on some arbitrary matrix space.

Let $\mathbf E_3$ be the elementary row 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 rows $i$ and $j$ transposed.

By the definition of a determinant:

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

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

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

Hence the result.

$\blacksquare$