Is it right to assume such that:
By proof of theorem, \begin{align*} \det(A)\det(B) &= \det(E_1E_2\dots E_k)\det(B)\\ &= \det(E_1E_2\dots E_kB)\\ &= \det(AB)\\ \end{align*}
Since both $A$ and $B$ are invertible and $\det(A)$ and $\det(B)$ both does not equal zero, one of them is a row echelon form of the other? For example,
A = $$ \left[ \begin{array}{cc} 0&3&3\\ -4&4&-1\\ 4&2&5 \end{array} \right] $$
B = $$ \left[ \begin{array}{cc} 4&2&5\\ 0&6&4\\ 0&0&1 \end{array} \right] $$
$B$ is a row echelon form of $A$, hence $B$ and $A$ are row equivalent since $\det(A)\det(B) \ne 0$
Is there a way to explain this concept in terms of elementary matrices?