Existence of Inverse Elementary Row Operation/Exchange Rows
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.
$\blacksquare$