Basic arithmetic with matrices We have matrices. 
$$A=\begin{bmatrix}
-2 & 0 \\
-5 & 6 \\
\end{bmatrix}
B^{-1}=\begin{bmatrix}
-7 & 8 \\
2 & -8 \\
\end{bmatrix}
C=\begin{bmatrix}
-15 & -2 \\
-8 & -14 \\
\end{bmatrix}
$$
We need to solve matrix $X$ from equation:
$$A^{-1}XB-C=0$$
$$X=AB^{-1}+C$$
$$X=\begin{bmatrix}
-2 & 0 \\
-5 & 6 \\
\end{bmatrix}
\begin{bmatrix}
-7 & 8 \\
2 & -8 \\
\end{bmatrix}
+
\begin{bmatrix}
-15 & -2 \\
-8 & -14 \\
\end{bmatrix}
$$
$$
X=\begin{bmatrix}
-1 & -18 \\
39 & -102 \\
\end{bmatrix}
$$
This doesn't seem to be correct solutions to this equations ?
 A: As point out by you that the order matters:
Pre-multiply and post-multiply are different.
Starting from
$${A^{ - 1}}XB - C = 0$$
Add $C$ on both sides
$${A^{ - 1}}XB = C$$
Now post-multiply by $B^{-1}$ on both sides
$${A^{ - 1}}X{B}{B^{ - 1}} = C{B^{ - 1}}$$
$${A^{ - 1}}X{I} = C{B^{ - 1}}$$
$${A^{ - 1}}X = C{B^{ - 1}}$$
Now pre-multiply by $A$ on both sides to get the final form
$${A}{A^{ - 1}}X = AC{B^{ - 1}}$$
$$X = AC{B^{ - 1}}$$
A: The equation
$X=AB^{-1}+C \tag 1$
does not follow from
$A^{-1}XB-C= 0; \tag 2$ 
instead, we have
$A^{-1}XB = C, \tag 3$
$XB = AC,\tag 4$
$X = ACB^{-1}; \tag 5$
if we now perform the indicated matrix arithmetic we arrive at
$X = \begin{bmatrix} -2 & 0 \\ -5 & 6 \end{bmatrix} \begin{bmatrix} -15 & -2\\-8 & -14 \end{bmatrix} \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}$
$= \begin{bmatrix} 30 & 4 \\ 27 & -74 \end{bmatrix}\begin{bmatrix} -7 & 8 \\ 2 & -8 \\ \end{bmatrix} = \begin{bmatrix} -202 & 208\\ -337 & 808 \\ \end{bmatrix}. \tag 6$
As a check of the above, we see that (1) implies
$A^{-1}XB = A^{-1}(AB^{-1}+C)B = A^{-1}AB^{-1}B + A^{-1}CB = I + A^{-1}CB \ne C \tag 7$ 
in general.
