Question

Solve for X, assuming all matrices involved are invertible: ABXA−1B−1 = I + A.

Answer 1

We have ABXA^-1 B^-1 = I + A.

On multiplying to the right by BA, we get (ABXA^-1 B^-1) BA = (I+A)BA or, (ABXA^-1)(B^-1 B)A = BA+ABA

or, (ABXA^-1)IA = BA+ABA or, (ABXA^-1)A = BA+ABA or, ABX(A^-1A) = BA+ABA or, ABXI = BA+ABA or, ABX = BA+ABA.

Now, on multiplying to the left by B^-1 A^-1, we get (B^-1 A^-1)ABX = (B^-1 A^-1)(BA+ABA)

or, B^-1 (A^-1A)BX = (B^-1 A^-1)(BA)+ (B^-1 A^-1)(ABA)

or, B^-1 IBX = (B^-1 A^-1)(BA)+ B^-1 (A^-1A)BA or, (B^-1B)X = (B^-1 A^-1)(BA)+ B^-1 IBA or, IX = (B^-1 A^-1)(BA)+ B^-1 BA

or, X = (B^-1 A^-1)(BA)+ IA   

or, X = (B^-1 A^-1)(BA)+A.