User:Zahlenspieler/Sandbox

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

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 $\operatorname{GL}\left({n, \R}\right)$.

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

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

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

by Definition:Inverse Matrix, we have:
 * $\displaystyle \mathbf{A}^\left({-1}\right) =\mathbf M$.

The rest follows from:
 * $\displaystyle \mathbf A^\left({-1}\right) \mathbf I=\mathbf A^\left({-1}\right)$.

Hence the result.