Yet another matrix equation 
Solve the following matrix equation for $X$.
$$\left[\begin{array}{cc}
5 &-8\cr
8 &1
\end{array}\right] X + \left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] = \left[\begin{array}{cc}
-1 &4\cr
-3 &-1
\end{array}\right] X$$

Please give me some hint to do this question. Thanks.

This is a linear equation in four variables. 
Let $X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}$.
Consider each entry of the RHS and LHS as a seperate equation:
$$\begin{align}
5x_{11} - 8x_{21} + 6 &= -1x_{11}+4x_{12} & \text{(top left entry)}\\
5x_{12} - 8x_{22} + 6 &= -1x_{12}+4x_{22} & \text{(top right entry)}\\
8x_{11} - 1x_{21} + 3 &= -3x_{11}-1x_{12} & \text{(bottom left entry)}\\
8x_{12} - 1x_{22} + 5 &= -3x_{12}-1x_{22} & \text{(bottom right entry)}\\
\end{align}$$
Now you get a linear system of four equations wich can be solved using the usual methods.

As an alternative you can directly use matrix operations and solve
$$\left[\begin{array}{cc}
5 &-8\cr
8 &1
\end{array}\right] X + \left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] = \left[\begin{array}{cc}
-1 &4\cr
-3 &-1
\end{array}\right] X$$
iff
$$\left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] = \left[\begin{array}{cc}
-1 &4\cr
-3 &-1
\end{array}\right] X - \left[\begin{array}{cc}
5 &-8\cr
8 &1
\end{array}\right] X  =\left[\begin{array}{cc}
-6 &12\cr
-11 &-2
\end{array}\right]X $$
iff 
$$\left[\begin{array}{cc}
-6 &12\cr
-11 &-2
\end{array}\right]^{-1}\left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] =X $$
(Note that the matrix that we are trying to invert is actually invertible since its determinant is $(-6)(-2)-(-11)12 = 144 \neq 0$.)
iff
$$X = \frac{1}{12}\left[\begin{array}{cc}
-4 & -6\cr
4 &3
\end{array}\right] $$
