Question Appendix.7: Using Proof by Contradiction in Linear Algebra Let A and B b...

Assume that neither A or B is singular. Because you know that a matrix is singular if and only if its determinant is zero, $det (A)$ and $det(B)$ are both nonzero real numbers. By Theorem 3.5, $det (AB) = det (A)det (B)$. So, $det (AB)$ is not zero because it is a product of two nonzero real numbers. But this contradicts that AB is a singular matrix. So, you can conclude that the assumption was wrong and that either A or B is singular.