Elementary operations on matrices 
$$A \cdot
\begin{bmatrix}
1 & 3 & 4\\
3 & -1 & 5\\
-2 & 4 & -3\\
\end{bmatrix}
=
\begin{bmatrix}
3 & -1 & 5\\
1 & 3 & 4\\
4 & -8 & 6\\
\end{bmatrix}
$$ Find the $3 \times 3$ matrix $A$.

According to my textbook, the question requires elementary row operations on the given matrices.
I read somewhere that for an equation of the form $AB=X$ ,we can apply elementary row operation on $A$ and $X$ only. I don't know why do these contradict. Where am I wrong?

Since
$$\det \begin{bmatrix} 1 & 3 & 4\\ 3 & -1 & 5\\ -2 & 4 & -3\end{bmatrix} = 20 \neq 0$$
we can right-multiply both sides of the linear matrix equation by elementary matrices until we obtain
$$\mathrm A = \begin{bmatrix} 3 & -1 & 5\\ 1 & 3 & 4\\ 4 & -8 & 6\end{bmatrix} \begin{bmatrix} 1 & 3 & 4\\ 3 & -1 & 5\\ -2 & 4 & -3\end{bmatrix}^{-1}$$
We would be doing elementary column operations. If you must do elementary row operations, then do transpose both sides of the linear matrix equation, then do left-multiply both sides by elementary matrices, obtain $\mathrm A^{\top}$ and then transpose to obtain $\mathrm A$.
