Solving for X in a simple matrix equation system. I am trying to solve for X in this simple matrix equation system:
$$\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix} - X\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix} = E $$ where $E$ is the identity matrix.
If I multiply $X$ with $\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix}$ I get the following system:
$$\begin{bmatrix}5x_1 & -1x_2\\6x_3 & -4x_4\\\end{bmatrix}$$
By subtracting this from $\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix}$ I get $\begin{bmatrix}7 - 5x_1 & 7 + 1x_2\\2 - 6x_3 & 4 + 4x_4\\\end{bmatrix} = \begin{bmatrix}1 & 0\\0 & 1\\\end{bmatrix}$
Which gives me:
$7-5x_1 = 1$
$7+1x_2 = 0$
$2-6x_3 = 0$
$4+4x_4 = 1$
These are not the correct answers, can anyone help me out here?
Thank you!

Lets set your equation to look like $C-XA = E$, where $C$ is the constant matrix and $A$ is your coefficient matrix.
first, isolate your matrix $AX$ by moving the constant matrix to the other side of he equation. manipulate the equation to give you $AX$ as positive. then multiply the matrix $A$ by its inverse.   Notice that this is a right hand multiplication.   Now you are left with your $X$ vector times the identity on the left.
Now you have
$$EX = (C-E)A^{-1}$$
You can combine $C$ and $E$ first, or distribute $A^{-1}$ through, it doesn't matter.  Just remember that you are doing a right hand multiplication, not left hand.  This is important because multiplication is not communicative in matrices.
