Prove that if A is an n times n inverible matrix then detA1

Prove that if A is an n times n inverible matrix, then det(A^-1) = 1/det(A).

Solution

Let A be a matrix of size n x n.

Suppose A is an invertible matrix then A. A^-1 = A^-1. A= I where I is an identity matrix of size n

Det (A . A^-1 ) = Det( A^-1. A )=det( I )

Det (A ).De t( A^-1 ) = Det( A^-1 ).Det ( A ) = 1

Therefore, Det ( A^-1 ) = 1/ Det( A )