Existence of Inverse Elementary Row Operation/Exchange Rows

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

\((\text {ERO} 3)\)   $:$   \(\ds r_k \leftrightarrow r_l \)    Exchange rows $k$ and $l$      


Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.


Then $e'$ is the elementary row operation:

$e' := r_k \leftrightarrow r_l$

That is:

$e' = e$


Proof

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 $\mathbf A = \mathbf A''$.


Let $\map e {\mathbf A}$ be the elementary row operation:

$e := r_k \leftrightarrow r_l$

Thus we have:

\(\ds r'_k\) \(=\) \(\ds r_l\)
\(\, \ds \text {and} \, \) \(\ds r'_l\) \(=\) \(\ds r_k\)


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:

\(\ds r''_k\) \(=\) \(\ds r'_l\)
\(\, \ds \text {and} \, \) \(\ds r''_l\) \(=\) \(\ds r'_k\)
\(\ds \leadsto \ \ \) \(\ds r''_k\) \(=\) \(\ds r_k\)
\(\, \ds \text {and} \, \) \(\ds r''_l\) \(=\) \(\ds r_l\)
\(\ds \leadsto \ \ \) \(\ds \mathbf A''\) \(=\) \(\ds \mathbf A\)

It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.

$\blacksquare$