Row Operation has Inverse
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 $\Gamma$ be a row operation which transforms $\mathbf A$ to a new matrix $\mathbf B \in \map \MM {m, n}$.
Then there exists another row operation $\Gamma'$ which transforms $\mathbf B$ back to $\mathbf A$.
Proof
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary row operations that compose $\Gamma$.
Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary row matrices.
From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have:
- $\mathbf R \mathbf A = \mathbf B$
where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
- $\mathbf R = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$
By Elementary Row Matrix is Invertible, each of $\mathbf E_i$ is invertible.
By Product of Matrices is Invertible iff Matrices are Invertible, it follows that $\mathbf R$ is likewise invertible.
Thus $\mathbf R$ has an inverse $\mathbf R^{-1}$.
Hence:
| \(\ds \mathbf R \mathbf A\) | \(=\) | \(\ds \mathbf B\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \mathbf R^{-1} \mathbf R \mathbf A\) | \(=\) | \(\ds \mathbf R^{-1} \mathbf B\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \mathbf A\) | \(=\) | \(\ds \mathbf R^{-1} \mathbf B\) |
We have:
| \(\ds \mathbf R^{-1}\) | \(=\) | \(\ds \paren {\mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1}^{-1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds {\mathbf E_1}^{-1} {\mathbf E_2}^{-1} \dotsb {\mathbf E_{k - 1} }^{-1} {\mathbf E_k}^{-1}\) | Inverse of Matrix Product |
From Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse, each of ${\mathbf E_i}^{-1}$ is the elementary row matrix corresponding to the inverse $e'_i$ of the corresponding elementary row operation $e_i$.
Let $\Gamma'$ be the row operation composed of the finite sequence of elementary row operations $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$.
Thus $\Gamma'$ is a row operation which transforms $\mathbf B$ into $\mathbf A$.
Hence the result.
$\blacksquare$
Also see
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.15 \ \text {(a)}$