Prove that applying Gauss-Jordan elimination to a square matrix augmented by identity matrix will yield the inverse matrix. I tried using induction to prove that $[A|I]$ will yield $[I|inverse(A)]$ by applying gauss-Jordan elimination on $[A|I]$ but couldnt prove it, though it works for several numerical base cases.
Can someone help me out?
 A: Recall that you may consider the multiplication of two matrices $A$ and $B$ that way: Multiply $A$ with the first column of $B$, then with the second one and so on.
Given $A$, you want to find a matrix $B$ such that $AB=I$.  Now $A$, multiplied by the first column $b_1$ of $B$, must yield the first column vector of $I$ which is the unit vector $e_1$, hence the first column of $B$ is the solution to $Ab_1=e_1$.  That system can be solved by Gauss-Jordan.  Similarly the second column $b_2$ of $B$ multiplied by $A$ must be the second column vector of $I$ which is $e_2$; hence $b_2$ is the solution of $Ab_2=e_2$ and so on.
Now all you have to do is solve simultaneously those equations; that's what's happening in $[A|I]$ transferring to $[I|B]$.
A: Suppose that elimination on $[A|I]$ produces $[I|E]$. Then there is a sequence of elementary matrices $E_1,\dots,E_k$ such that $E=E_k\dots E_1I$ and $E_k\dots E_1A=EA=I$, proving $E$ to be a left-inverse of $A$. Since elementary matrices are invertible, $E$ has a two-sided inverse $E^{-1}=E_1^{-1}\dots E_k^{-1}$. It follows that
\begin{align}
A=IA=(E^{-1}E)A=E^{-1}(EA)=E^{-1}I=E^{-1}\,.
\end{align}
