If A and B are invertible square matrices of the same order then (AB)-1=?
(AB)(AB)-1 = I
A-1(AB)(AB)-1 = IA-1
(A-1A)B (AB)-1=A-1
IB(AB)-1 = A-1
B(AB)-1 = A-1
B-1B(AB)-1 = B-1 A-1
I (AB)-1 = B-1A-1
(AB)-1 = B-1A-1