I'm facing a problem with solving the following equation: $$ B^{-1}XA = -B^{-1}X+2E $$ Where: $$ A = \begin{bmatrix}2&0\\0&3\end{bmatrix}, B = \begin{bmatrix}2&1\\2&2\end{bmatrix} $$ And E is the appropriate identity matrix.
I tried multiple ways (by first calculating the inverse of B, multiplying it with A, adding $-B^{-1}X$ to both sides so I'd get $B^{-1}*X*A+B^{-1}X = 2E$, then after getting the inverse of B multiplied with A, I also added inverse of B to it, giving me $X(B^{-1}*A+B^{-1}) = 2E$. So I inverted the $(B^{-1}*A+B^{-1})$ to satisfy $A^{-1}*A = E, E*A = A$, which ended up with $X = 2E* (B^{-1}*A+B^{-1})^{-1}$
I might've done something wrong, maybe the calculations aren't allowed this way or something, anyhow the matrix which I got for A, didn't satisfy the equation.
Best regards!
\begin{align} B^{-1}XA = -B^{-1}X+2I & \\ \implies& BB^{-1}XA = -BB^{-1}X+2B & \\ \implies& XA=-X+2B & \\ \implies& XA+X=2B & \\ \implies & X(A+I)=2B & \\ \implies & X(A+I)(A+I)^{-1}=2B(A+I)^{-1} & \\ \implies & X=2B(A+I)^{-1}. \end{align}