Why does the Gaussian-Jordan elimination work when finding the inverse matrix? In order to find the inverse matrix $A^{-1}$, one can apply Gaussian-Jordan elimination to the augmented matrix
$$(A \mid I)$$
to obtain
$$(I \mid C),$$
where $C$ is indeed $A^{-1}$. However, I fail to see why this actually works, and reading this answer didn't really clear things up for me.

You want to find a matrix $B$ such that $BA = I$. $B$ can be written as a product of elementary matrices iff it is invertible. Hence we attempt to obtain $B$ by left-multiplying $A$ by elementary matrices until it becomes $I$. All we have to do is to keep track of the product of those elementary matrices. But that is exactly what we are doing when we left-multiply $I$ by those same elementary matrices in the same order. This is what is happening with the augmented matrix.
