Proof of Theorem DRMM

This proof is constructed in two cases. First, suppose that $A$ is singular. Then $\detname{A}=0$ by Theorem SMZD. By the contrapositive of Theorem NPNT, $AB$ is singular as well. So by a second application of Theorem SMZD, $\detname{AB}=0$. Putting it all together \begin{align*} \detname{AB}=0=0\,\detname{B}=\detname{A}\detname{B} \end{align*} as desired.

For the second case, suppose that $A$ is nonsingular. By Theorem NMPEM there are elementary matrices $E_{1},\,E_{2},\,E_{3},\,\dots,\,E_{s}$ such that $A=E_1E_2E_3\dots E_s$. Then \begin{align*} \detname{AB} &= \detname{E_1E_2E_3\dots E_{s}B}\\ &= \detname{E_1}\detname{E_2}\detname{E_3}\dots\detname{E_s}\detname{B} &&\knowl{./knowls/theorem.DEMMM.knowl}{\text{Theorem DEMMM}}\\ &= \detname{E_1E_2E_3\dots E_s}\detname{B} &&\knowl{./knowls/theorem.DEMMM.knowl}{\text{Theorem DEMMM}}\\ &= \detname{A}\detname{B} \end{align*}