Proof of Theorem DEMMM

The proof procedes in three parts, one for each type of elementary matrix, with each part very similar to the other two.

First, let $B$ be the matrix obtained from $A$ by swapping rows $i$ and $j$, \begin{align*} \detname{\elemswap{i}{j}A} &=\detname{B}&&\knowl{./knowls/theorem.EMDRO.knowl}{\text{Theorem EMDRO}}\\ &=-\detname{A}&&\knowl{./knowls/theorem.DRCS.knowl}{\text{Theorem DRCS}}\\ &=\detname{\elemswap{i}{j}}\detname{A}&&\knowl{./knowls/theorem.DEM.knowl}{\text{Theorem DEM}} \end{align*}

Second, let $B$ be the matrix obtained from $A$ by multiplying row $i$ by $\alpha$, \begin{align*} \detname{\elemmult{\alpha}{i}A} &=\detname{B}&&\knowl{./knowls/theorem.EMDRO.knowl}{\text{Theorem EMDRO}}\\ &=\alpha\detname{A}&&\knowl{./knowls/theorem.DRCM.knowl}{\text{Theorem DRCM}}\\ &=\detname{\elemmult{\alpha}{i}}\detname{A}&&\knowl{./knowls/theorem.DEM.knowl}{\text{Theorem DEM}} \end{align*}

Third, let $B$ be the matrix obtained from $A$ by multiplying row $i$ by $\alpha$ and adding to row $j$, \begin{align*} \detname{\elemadd{\alpha}{i}{j}A} &=\detname{B}&&\knowl{./knowls/theorem.EMDRO.knowl}{\text{Theorem EMDRO}}\\ &=\detname{A}&&\knowl{./knowls/theorem.DRCMA.knowl}{\text{Theorem DRCMA}}\\ &=\detname{\elemadd{\alpha}{i}{j}}\detname{A}&&\knowl{./knowls/theorem.DEM.knowl}{\text{Theorem DEM}} \end{align*}

Since the desired result holds for each variety of elementary matrix individually, we are done.