Is the inverse of any invertible matrices still invertible W

Solution

1. If A is invertible, then det(A) 0 . Further , since det(A^-1) = 1/det(A), hence det(A^-1) 0 so that A^-1 is also invertible. Thus, for invertible matrix A, there exists an invertible matrix A^-1 such that (AA^-1) =(A^-1A) = I. Then, by the definition of the inverse, (A^-1)^-1 = A.

2. Let A,B be two nxn invertible matrices. Then det(A) 0 and det(B) 0 so that det(AB) 0. Hence AB is invertible. Let us denote B^-1A^-1 by C . We will show that (AB)C=C(AB)=I. On substituting B^-1A^-1 for C, we get:

    (AB)C = (AB)(B^-1A^-1) = ABB^-1A^-1 = A(BB^-1)A^-1 = AIA^-1 = AA^-1 = I.

    Similarly C(AB) = (B^-1A^-1)(AB) = B^-1A^-1AB = B^-1( A^-1A)B = B^-1 IB = B^-1B = I. Thus, by the definition of inverse, C = B^-1A^-1 is the inverse of AB