Proposition 2.3.1 (Solving equations with matrix inverse). If $A$ has an inverse $A^{-1}$, then for any $\vec b$ the equation $A\vec x = \vec b$ has a unique solution, namely $\vec x = A^{-1}\vec b$.
If $A$ is a $n\times n$ matrix, and you construct the $n\times 2n$ augmented matrix $[A|I]$ and row reduce it, then either:
Example. The matrix $A = \begin{bmatrix}2&1&3\\1&-1&1\\1&1&2\end{bmatrix}$ has inverse $A^{-1} = \begin{bmatrix}3&-1&-4\\1&-1&-1\\-2&1&3\end{bmatrix}$, because
$\begin{bmatrix}2&1&3&1&0&0\\1&-1&1&0&1&0\\1&1&2&0&0&1\end{bmatrix}$ row reduces to $\begin{bmatrix}1&0&0&3&-1&-4\\0&1&0&1&-1&-1\\0&0&1&-2&1&3\end{bmatrix}$.
Each one of the three row operations on a matrix $A$ can be represnted by multiplication on the left by an elementary matrix. All of these matricies are square.
Writing row operations as matrices lets us use properties of matrices to our advantage.
If $A$ is invertible, then $A^{-1}$ is equal to a product of elementary matrices such that
$I = E_k\dots E_2E_1A$.