I had the equation $AX + B = CX$
which I rearranged to be X = $(C-A)^{-1}$ B
BUT I'm not quite sure how to go about solving $(C-A)^{-1}$. I have a rule written down in my notes that $(A+B)^{-1}$ is not equal to $A^{-1} + B^{-1}$ and I'm assuming that's the same for subtraction. Could some one please help me out? Thank you!!
There is no nice form in general for $(C-A)^{-1}$, but since they're just $2\times 2$ matrics, you can calculate $C-A$ by hand, and the calculate the inverse of that by hand, remembering that $$ \begin{pmatrix}a & b\\c & d\end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix}d & -b\\-c & a\end{pmatrix} \,,$$ then multiply that result by $B$ by hand. This calculation is not too arduous.