Solving for x with matrices $$
        \begin{bmatrix}
        -5 & -9  \\
        -6 & -2  \\
        \end{bmatrix}   X +   \begin{bmatrix}
        -3 & 2  \\
        -9 & -3  \\
        \end{bmatrix}   =  \begin{bmatrix}
        -9 & -7  \\
        -2 & -3  \\
        \end{bmatrix}   X
$$ 
I am asked to solve for matrix X
I combine the x to the right side and get
$$
        \begin{bmatrix}
        -14 & -16  \\
        -8 & -6  \\
        \end{bmatrix}   X
$$  
Then I move the other matrix to the right and get
$$
        \begin{bmatrix}
        -14 & -16  \\
        -8 & -6  \\
        \end{bmatrix}   X = \begin{bmatrix}
        3 & -2  \\
        9 &  3  \\
        \end{bmatrix}
$$  
I then proceed to find the inverse of the x coefficient matrix which comes out to be \begin{bmatrix}
        6/44 & -16/44  \\
        -8/44 & 14/44  \\
        \end{bmatrix}
I know this is the correct inverse because when I multiply it by the original x coefficient matrix I get the identity matrix.
Now as far as I'm aware, all that's left is multiplying the inverse by the right side, but when I do that, my homework system tells me that I'm wrong. Where is the mistake? This is the answer that I get
\begin{bmatrix}
        -63/22 & -15/11  \\
        51/22 & 29/22  \\
        \end{bmatrix} 
 A: You have a sign error, as
$$ \left( \begin{pmatrix} -5 & -9 \\ -6 & -2 \end{pmatrix} - \begin{pmatrix} -9 & -7 \\ -2 & -3 \end{pmatrix} \right) X = \begin{pmatrix} 4 & -2 \\ -4 & 1 \end{pmatrix}X = \begin{pmatrix} 3 & -2 \\ 9 & 3 \end{pmatrix}.  $$
A: The +/- signs are wrong on various places. RHS should just be the matrix displayed, because it already has - sign.
Also with the matrix before x you made a sign error
A: You have made a few mistakes in signs while reversing the sides.
You should have approached the problem as follows:
$$
        \begin{bmatrix}
        -5 & -9  \\
        -6 & -2  \\
        \end{bmatrix}   x -   \begin{bmatrix}
        -3 & 2  \\
        -9 & -3  \\
        \end{bmatrix}   =  \begin{bmatrix}
        -9 & -7  \\
        -2 & -3  \\
        \end{bmatrix}   x
$$ 
$$
        \begin{bmatrix}
        -5x & -9x  \\
        -6x & -2x  \\
        \end{bmatrix}    -   \begin{bmatrix}
        -3 & 2  \\
        -9 & -3  \\
        \end{bmatrix}   =  \begin{bmatrix}
        -9x & -7x  \\
        -2x & -3x  \\
        \end{bmatrix}   
$$ 
$$
        \begin{bmatrix}
        -5x+3 & -9x-2  \\
        -6x+9 & -2x+3  \\
        \end{bmatrix}  =  \begin{bmatrix}
        -9x & -7x  \\
        -2x & -3x  \\
        \end{bmatrix}   
$$ 
By the equality of the matrices, we can say that the corresponding elements are the same. 
But there is no $x \in \mathbb{R}$ which satisfies the above equality.
A: You have formally
$$ AX +B = CX$$
You should proceed as follows
$$ \begin{array}{rll}
 AX + B &=& CX \\
  B &=& CX - AX \\
  B &=& (C- A)X \\
(C-A)^{-1}  B &=& (C-A)^{-1} (C- A)X \\
(C-A)^{-1}  B &=& X \\
 \end{array} $$
Now $$ C-A = -\begin{bmatrix} 9 & 7 \\ 2 &3  \end{bmatrix}
 - (-\begin{bmatrix} 5 & 9 \\ 6 &2  \end{bmatrix})
= \begin{bmatrix} -4 & 2 \\ 4 & -1  \end{bmatrix}  $$
Then $$ (C-A)^{-1} = -\begin{bmatrix} 1/4 & 1/2 \\ 1 &1  \end{bmatrix} $$
 and  by $ (C-A)^{-1} B=X$ the result  

  $$ (C-A)^{-1} B = - \begin{bmatrix} 21/4 & 1 \\ 12 &1  \end{bmatrix} = X $$

