Knowing that $A=\begin{pmatrix} -1 &2 \\ 1&1 \end{pmatrix},B=\begin{pmatrix} 1 & 2\\ 0&1 \end{pmatrix},C=\begin{pmatrix} 2 &2 \\ 3 &1 \end{pmatrix}$ solve the system
$$\left\{\begin{matrix} AX+CY=BAC\\ CX-BY=CAB \end{matrix}\right.$$ So guys how do I solve this. I know how to find the inverse of matrix and I guess I should use it can you show me how to start and what to do after?
$\begin{matrix} C^{-1}AX+Y=C^{-1}BAC\\ B^{-1}CX-Y=B^{-1}CAB \end{matrix}$
Great, add the two together.
$(C^{-1}A+B^{-1}C)X=C^{-1}BAC + B^{-1}CAB\\ X=(C^{-1}A+B^{-1}C)^{-1}C^{-1}BAC + B^{-1}CAB $
It may feel a little labor intensive to calculate that all out, but it isn't really that big of a killer.
$Y = C^{-1}BAC - C^{-1}AX$