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!
 A: $$\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix} - X\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix} = I $$ where $I$ is the identity matrix.
Hint: reconsider what multiplication by $X$ will look like:
$X$ will be a $2\times 2$ matrix, if matrix multiplication and addition is to be defined for this equation.
So if $X = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}$, then $$X\begin{bmatrix}5 & -1\\6 & -4 \end{bmatrix} = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & -4\end{bmatrix} = \begin{bmatrix}5x_1+6x_2&-x_1-4x_2\\5x_3+6x_4&-x_3-4x_4\end{bmatrix}$$
A: Since $\begin{pmatrix}7&7\\2&4\end{pmatrix}-X\begin{pmatrix}5&-1\\6&-4\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, we obtain:
$\begin{pmatrix}6&7\\2&3\end{pmatrix}=\begin{pmatrix}5x_1+6x_2&-x_1-4x_2\\5x_3+6x_4&-x_3-4x_4\end{pmatrix}$, where $X=\begin{pmatrix}x_1&x_2\\x_3&x_4\end{pmatrix}$.
Now you can multiply both sides of the equation by $\frac{1}{-14}\begin{pmatrix}-4&1\\-6&5\end{pmatrix}$ =(inverse of $\begin{pmatrix}5&-1\\6&-4\end{pmatrix}$), to find:
$X=\frac{1}{-14}\begin{pmatrix}6&7\\2&3\end{pmatrix}\begin{pmatrix}-4&1\\-6&5\end{pmatrix}=\frac{1}{-14}\begin{pmatrix}-66&41\\-26&17\end{pmatrix}$.
Hope this helps.
A: 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.
