# Exchange of Rows as Sequence of Other Elementary Row Operations

## Theorem

Let $\mathbf A$ be an $m \times n$ matrix.

Let $i, j \in \closedint 1 m: i \ne j$

Let $r_k$ denote the $k$th row of $\mathbf A$ for $1 \le k \le m$:

$r_k = \begin {pmatrix} a_{k 1} & a_{k 2} & \cdots & a_{k n} \end {pmatrix}$

Let $e$ be the elementary row operation acting on $\mathbf A$ as:

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

Then $e$ can be expressed as a finite sequence of exactly $4$ instances of the other two elementary row operations.

 $(\text {ERO} 1)$ $:$ $\ds r_i \to \lambda r_i$ For some $\lambda \in K_{\ne 0}$, multiply row $i$ by $\lambda$ $(\text {ERO} 2)$ $:$ $\ds r_i \to r_i + \lambda r_j$ For some $\lambda \in K$, add $\lambda$ times row $j$ to row $i$

## Proof

In the below:

$r_i$ denotes the initial state of row $i$
$r_j$ denotes the initial state of row $j$
$r_i'$ denotes the state of row $i$ after having had the latest elementary row operation applied
$r_j'$ denotes the state of row $j$ after having had the latest elementary row operation applied.

$(1)$: Apply $\text {ERO} 2$ to row $j$ for $\lambda = 1$:

$r_j \to r_j + r_i$

After this operation:

 $\ds r_i'$ $=$ $\ds r_i$ $\ds r_j'$ $=$ $\ds r_i + r_j$

$\Box$

$(2)$: Apply $\text {ERO} 2$ to row $i$ for $\lambda = -1$:

$r_i \to r_i + \paren {-r_j}$

After this operation:

 $\ds r_i'$ $=$ $\ds r_i - \paren {r_i + r_j}$ $\ds$ $=$ $\ds -r_j$ $\ds r_j'$ $=$ $\ds r_i + r_j$

$\Box$

$(3)$: Apply $\text {ERO} 2$ to row $j$ for $\lambda = 1$:

$r_j \to r_j + r_i$

After this operation:

 $\ds r_i'$ $=$ $\ds -r_j$ $\ds r_j'$ $=$ $\ds r_i + r_j - r_j$ $\ds$ $=$ $\ds r_i$

$\Box$

$(4)$: Apply $\text {ERO} 1$ to row $i$ for $\lambda = -1$:

$r_i \to -r_i$

After this operation:

 $\ds r_i'$ $=$ $\ds -\paren {-r_j}$ $\ds$ $=$ $\ds r_j$ $\ds r_j'$ $=$ $\ds r_i$

$\Box$

Thus, after all the $4$ elementary row operations have been applied, we have:

 $\ds r_i'$ $=$ $\ds r_j$ $\ds r_j'$ $=$ $\ds r_i$

Hence the result.

$\blacksquare$