Let A and B be nSolutionBy Rule if none of the eigenvalues o

Let A and B be n

Solution

By Rule if none of the eigenvalues of A is equal to 1, then I?A is invertible.

So, let I-A = B

Multiply both sides with X

X(I-A) = XB

X-XA = B (Since matrix B is invertible)

X= XA + B

Hence XA+B=X

Solution

XA -X = B

X(A-1)=B

X = B * (A-I)^(-1)

So it has unique solution.