Existence of Inverse Elementary Row Operation

Theorem
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.

Let $\mathbf A \in \map \MM {m, n}$ be a matrix.

Let $\map e {\mathbf A}$ be an elementary row operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.

Then there exists another elementary row operation $\map {e'} {\mathbf A'}$ which transforms $\mathbf A'$ back to $\mathbf A$.

Proof
Let us take each type of elementary row operation in turn.

For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new matrix $\mathbf A'' \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.

In the below, let:
 * $r_k$ denote row $k$ of $\mathbf A$
 * $r'_k$ denote row $k$ of $\mathbf A'$
 * $r_k$ denote row $k$ of $\mathbf A$

for arbitrary $k$ such that $1 \le k \le m$.

By definition of elementary row operation:
 * only the row or rows directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$

and similarly:
 * only the row or rows directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A''$.

Hence it is understood that in the following, only those rows directly affected will be under consideration when showing that $\mathbf A = \mathbf A''$.

$\text {ERO} 1$: Scalar Product of Row
Let $\map e {\mathbf A}$ be the elementary row operation:


 * $e := r_k \to \lambda r_k$

where $\lambda \ne 0$.

Then $r'_k$ is such that:
 * $\forall a'_{k i} \in r'_k: a'_{k i} = \lambda a_{k i}$

Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$:
 * $e' := r_k \to \dfrac 1 \lambda r_k$

Because it is stipulated in the definition of an elementary row operation that $\lambda \ne 0$, it follows by definition of a field that $\dfrac 1 \lambda$ exists.

Hence $e'$ is defined.

So applying $e'$ to $\mathbf A'$ we get:

$\text {ERO} 2$: Add Scalar Product of Row to Another
Let $\map e {\mathbf A}$ be the elementary row operation:


 * $e := r_k \to r_k + \lambda r_l$

Then $r'_k$ is such that:
 * $\forall a'_{k i} \in r'_k: a'_{k i} = a_{k i} + \lambda a_{l i}$

Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$:
 * $e' := r'_k \to r'_k - \lambda r'_l$

Applying $e'$ to $\mathbf A'$ we get:

$\text {ERO} 3$: Exchange Rows
Let $\map e {\mathbf A}$ be the elementary row operation:


 * $e := r_k \leftrightarrow r_l$

Thus we have:

Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$:
 * $e' := r'_k \leftrightarrow r'_l$

Applying $e'$ to $\mathbf A'$ we get:

Thus in all cases, for each elementary row operation which transforms $\mathbf A$ to $\mathbf A'$, we have constructed an elementary row operation which transforms $\mathbf A'$ to $\mathbf A$.

Hence the result.