Solve the given matrix equation for X. Simplify your answers as much as possible. (In the words of Albert Einstein, "Everything should be made as simple as possible, but not simpler.") Assume that all matrices are invertible. ABXA−1B−1 = I + A

Answer:[tex]X=B^{-1}A^{-1}BA+A[/tex]Step-by-step explanation:We are given that a  matrix equation [tex]ABXA^{-1}B^{-1}=I+A[/tex]We have to solve the  given matrix equation for X.Suppose all matrix are invertible.Left multiply by [tex]B^{-1}A^{-1}[/tex] on both sides then ,we get [tex]B^{-1}A^{-1}ABXA^{-1}B^{-1}=B^{-1}A^{-1}(I+A)[/tex][tex]B^{-1}BXA^{-1}B^{-1}=B^{-1}A^{-1}+B^{-1}A^{-1}A[/tex]    [tex]AA^{-1}=A^{-1}A=I[/tex] When A is invertible.[tex]XA^{-1}B^{-1}=B^{-1}A^{-1}+B^{-1}[/tex]     ([tex]B^{-1}B=BB^{-1}=I)[/tex]Right multiply by BA on both sides then we get [tex]XA^{-1}B^{-1}BA=B^{-1}A^{-1}BA+B^{-1}BA[/tex][tex]XA^{-1}A=B^{-1}A^{-1}BA+A[/tex][tex]XI=B^{-1}A^{-1}BA+A[/tex][tex]X=B^{-1}A^{-1}BA+A[/tex]        (XI=X)