# A proof involving invertible $n\times n$ matrices

I'm new to studying linear algebra... doing ok with the computation type questions but struggling through some of the proof questions. I'm just not sure how to get started and what format this type of proof should take. Thanks in advance for suggestions. :)

Let $A$ be an non singular $n\times n$ matrix and $B$ be an $n\times k$ matrix. Assume that RREF of $[A|B]$ is $[I|X]$. Prove that $X=(A^{-1})B$.

Denote by $Be_1, \dots, Be_k$ the columns of $B$. Since row operations preserve the solution space, it means that if the RREF of $[A|b]$ is $[I|y]$ then the linear system of equations $Ax = b$ is equivalent to the system $Ix = x = y$ and so has a unique solution $x = y$. In your case, if the RREF of $[A|B] = [A|Be_1,\dots,Be_k] = [I|X]$, this means that $A(Xe_i) = (Be_i)$ for all $1 \leq i \leq k$ and so $AX = B$. Multiplying both sides by $A^{-1}$ on the left, we get $X = A^{-1}B$.
Row reduction of $[A|B]$ can be thought of as left-multiplying both $A$ and $B$ by a sequence $E_1, \dots, E_m$ of elementary $n \times n$ matrices in succession (which performs the $m$ row operations), or, more succinctly, by the single matrix $E=E_m \cdots E_1$. Since the result is $[I|X]$, we must have $EA =I$ and $E B = X$. The first equation implies $E = A^{-1}$, so the second equation becomes $X=A^{-1} B$, as desired.