In SGTA 8, we showed that for m xm matrices A and B with det A 0 and det B 0, that det(AB) = (det A)(det B). Show that det(AB) = (det A)(det B) holds also when det A = 0 and/or det B = 0. That is, prove that det(AB) = 0 if det A =0 and/or det B= 0. You may wish to consider the two cases: (a) det B = 0 (for arbitrary A, allowing for the possibility det A = 0), (b) det A =0 (for arbitrary B, allowing for the possibility det B = 0). You may use any result from the Appendix or the SGTAS (we showed that det CT det C for a 3x3 matrix C but you may take as fact that this holds for any square matrix). Hint for Cie (a), consider the homogeneous equation (AB)x = 0, and use the result from SGTA 8, Section 2, w6.