Question

Show that if A is invertible, then det A-=1/det A.

What theorem should be used to examine the quantity det A-1? Choose the correct answer below.
A square matrix A is invertible if and only if det A
B. If A is an n times n matrix, then det AT = det A.
C. If A and B are n times n matrices, then det AB = (det A)(det B).
D. If one row of a square matrix A is multiplied by k to produce B, then det B=k (det A)

Answer 1

To show that if A is invertible, then det A^-1=1/det A, we can use the formula det AB = (det A)(det B) and the fact that A^-1 A = I, where I is the identity matrix. Multiplying both sides of A^-1 A = I by det A^-1, we get det A^-1 A = det I = 1. Using the formula det AB = (det A)(det B), we can rewrite det A^-1 A = det A^-1 (det A) = 1, which gives us det A^-1 = 1/det A.

The correct answer to the second question is A. If A is an n times n matrix, then det A. The theorem states that a square matrix A is invertible if and only if its determinant is nonzero. In other words, det A ≠ 0 if and only if A is invertible.