the equation given is $A=(AX+B)^{-1}$, I first multiply by $(AX+B)$ and this yields that $A(AX+B)=I$ I distribute A, $A^2X+AB=I \implies A^2X=I-AB \implies X=(A^2)^{-1}(I-AB)$ If I know try and solve what $(I-AB)$ is, lets say it is C and then compute $A^{-1}$ and compute $(A^{-1})^2$ and then finally compute $(A^{-1})^2 \cdot C$ I get the wrong answer. If I however continue to simplify from this point:$ X=(A^2)^{-1}(I-AB) \implies X=A^{-1}(A^{-1}-B)$ from here I now get the correct answer, how come?
$A=\big(\begin{smallmatrix} 3 & 2\\ 1 & 1 \end{smallmatrix}\big)$ and $B=\big(\begin{smallmatrix} 2 & 0\\ 1 & 3 \end{smallmatrix}\big)$ If I compute $(A^2)^{-1}(I-AB)$ I get $X=\big(\begin{smallmatrix} 0 & -30\\ 0 & 41 \end{smallmatrix}\big)$ and$ A^{-1}(A^{-1}-B) $ gives $ X=\big(\begin{smallmatrix} 3 & -2\\ -5 & 2 \end{smallmatrix}\big)$ which is the correct answer.
For $A^2$ you should have $$ \begin{bmatrix} 11 & 8 \\ 4 & 3 \\ \end{bmatrix} $$ and for $(A^2)^{-1}$ you should have $$ \begin{bmatrix} 3 & -8 \\ -4 & 11 \\ \end{bmatrix} $$ And you should have the correct answer either way.