User:Zahlenspieler/Sandbox

Theorem
Let $\mathbf A$ be a square matrix of order $n$ of the matrix space $\map {\mathbf M_n} \R$.

Let $\mathbf I$ be the unit matrix of order $n$.

Suppose there exists a sequence of elementary row operations that reduces $\mathbf A$ to $\mathbf I$.

Then $\mathbf A$ is invertible.

Futhermore, the same sequence, when performed on $\mathbf I$, results in the inverse of $\mathbf A$.

Proof
For ease of presentation, let $\breve{\mathbf X}$ be the inverse of $\mathbf X$.

We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations.

By repeated application of Elementary Row Operations as Matrix Multiplications, we can write this assertion as:

From Elementary Row Matrix is Invertible:
 * $\mathbf {E_1}, \ldots, \mathbf{E_t}$ are elements of the General Linear Group $\GL {n, \R}$.

By General Linear Group is Group, there exists a square matrix of order $n$ in $\GL {n, \R}$ such that:
 * $\mathbf {E_t} \cdot \mathbf {E_{t - 1} } \cdots \mathbf {E_1} = \mathbf M$

So the assertion can be rewritten as:
 * $\mathbf {MA} = \mathbf I$.

By Left or Right Inverse of Matrix is Inverse, it follows that:
 * $\mathbf {MA} = \mathbf I$

By Definition:Inverse Matrix, we have:
 * $\mathbf A^{-1} = \mathbf M$

The rest follows from:
 * $\mathbf A^{-1} \mathbf I = \mathbf A^{-1}$

Hence the result.