Prove that if A is an n times n invertible matrix the detA1

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

Solution

Assume A is an invertible matrix. That means A^(-1) exists.
A matrix multiplied by its inverse is equal to the identity matrix, I.

A * A^(-1) = I

Take the determinant of both sides,

det( A * A^(-1) ) = det(I)

The determinant of the identity matrix is 1.

det( A * A^(-1) ) = 1

Applying the stated determinant identity,

det(A) det(A^(-1)) = 1

Since determinant result in real numbers after finding the determinant, we can divide both sides by det(A). Note that det(A) is nonzero because A is invertible, so we needn\'t worry about the possibility of dividing by zero.

det( A^(-1) ) = 1/det(A)