Solve for $X$ in a simple $2\times2$ equation system. I posted a similar question recently but I still have problem with this problem and would appreciate any help!
$$\left[ \begin{array}{cc} 9 & -3\\ 5 & -5\end{array} \right] - X \left[ \begin{array}{cc} -9 & -2\\ 8 & 5\end{array} \right] = E$$ With $E$ i pressume they mean the identity matrix $\left[ \begin{array}{cc} 1 & 0\\ 0 & 1\end{array} \right]$.
How should I go on and solve this for the $2\times2$ matrix $X$? a full development so I can follow your solution would be very much appreciated!
Thank you kindly for you help!
 A: I will express it as equations, as I think that is easier (at least for beginners).
With 
$$X=\begin{pmatrix} x_{11} & x_{12}\\ x_{21} & x_{22} \\ \end{pmatrix}$$
At first we make the multiplication 
\begin{align*}
X \cdot \begin{pmatrix} -9 & -2\\ 8 & 3 \end{pmatrix}&=
  \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \cdot 
    \begin{pmatrix} -9 & -2 \\ 8 & 3 \end{pmatrix}\\
&= \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\
                    -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\
    \end{pmatrix}
\end{align*}
So our equation is 
$$\begin{pmatrix} 9 & -3 \\ 5 & -5 \end{pmatrix} - \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\
                    -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\
    \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$
Now we write it as a system of equations:
\begin{align*}
1&=9- (-9 x_{11} +8x_{12})\\
0&= -3-(-2x_{11} +3x_{12}) \\
0&=5-(-9x_{21} +8 x_{21}) \\
1&=-5 - (-2x_{21}+3x_{22}) \\ 
\end{align*}
If you need help solving this system tell me.
A: First of all rearrange to get X[-9 -2, 8 5] = [8 -3, 5 -6]. Then find the inverse of [-9 -2, 8 5], and multiply both sides of the equation by this inverse on the right, which will leave X = [8 -3, 5 -6][-9 -2, 8 5]^-1 as your solution.
A: Denote $B=\begin{pmatrix}-9&-2\\8&5\end{pmatrix}$. Then we have:
$$\begin{pmatrix}9&-3\\5&-5\end{pmatrix}-E=XB$$
Observe that $\det B=-45+16\neq0$ and hence $B$ is invertible. Multiplying by $B^{-1}$ we have:
$$X=\left(\begin{pmatrix}9&-3\\5&-5\end{pmatrix}-E\right)B^{-1}$$
Do you know how to find $B^{-1}$?
