I've been reading Hoffman Kunze, and I came across this theorem (theorem $9$) which has a long and tedious proof. I've been wondering wether there could be a more elegant proof.
Theorem 9. Let $e$ be an elementary row operation and let $E$ be the $m\times m$ elementary matrix $E = e(I)$. Then, for every $m \times n$ matrix $A$, $$e(A) = EA.$$
To be more explicit about what I mean by 'elegant' proof, I'm looking for a proof where we don't actually have to consider each of the 3 elementary row operations separately.
$$e(A)=e(IA)=e(I)A=EA.$$
In fact, in general, you can say
$$e(BA)=e(B)A.$$
(This is justified below. The intuition behind this is that applying a row operation to a matrix and then multiplying it with another matrix is the same as applying a row operation to their product.)
Alternatively, you can reduce the three cases in the proof you're given to scalar multiplication (which is obvious) and adding rows, since a row change is just a combination of these operations.