Every elementary row operation has a unique inverse

Dependencies:

Let $A$ be a matrix over a field. Then every elementary row operation on $A$ has a unique inverse which is also an elementary row operation.

This means that if applying $R$ on $A$ gives $B$ , then applying the inverse of $R$ (denoted as $R^{- 1}$ ) on $B$ gives $A$ .

Inverse of $⟨ i ⟩ \leftarrow c ⟨ i ⟩$ is $⟨ i ⟩ \leftarrow (c^{- 1}) ⟨ i ⟩$ .
Inverse of $⟨ i ⟩ \leftarrow ⟨ i ⟩ + c ⟨ j ⟩$ $(i \neq j)$ is $⟨ i ⟩ \leftarrow ⟨ i ⟩ - c ⟨ j ⟩$ .
$⟨ i ⟩ \leftrightarrow ⟨ j ⟩$ is its own inverse.

By the above means of finding inverse, it can be seen that the inverse of $R^{- 1}$ is $R$ .

Suppose $B$ is obtained from $A$ by applying the operation $R$ . Let $a_{i}$ be the $i^{th}$ row of $A$ and $b_{i}$ be the $i^{th}$ row of $B$ .

$⟨ p ⟩ \leftarrow c ⟨ p ⟩$ : $b_{i} = {\begin{cases} a_{i} & i \neq p \\ c a_{i} & i = p \end{cases} ⟹ a_{i} = {\begin{cases} b_{i} & i \neq p \\ (c^{- 1}) b_{i} & i = p \end{cases}$
$⟨ p ⟩ \leftarrow ⟨ p ⟩ + c ⟨ q ⟩$ where $p \neq q$ : $b_{i} = {\begin{cases} a_{i} & i \neq p \\ a_{p} + c a_{q} & i = p \end{cases} ⟹ a_{i} = {\begin{cases} b_{i} & i \neq p \\ b_{p} - c a_{q} & i = p \end{cases} = {\begin{cases} b_{i} & i \neq p \\ b_{p} - c b_{q} & i = p \end{cases}$
$⟨ p ⟩ \leftrightarrow ⟨ q ⟩$ : $b_{i} = {\begin{cases} a_{i} & i \neq p \land i \neq q \\ a_{q} & i = p \\ a_{p} & i = q \end{cases} ⟹ a_{i} = {\begin{cases} b_{i} & i \neq p \land i \neq q \\ b_{p} & i = q \\ b_{q} & i = p \end{cases}$