If $E_1,E_2,…E_q$ are the elementary matrices corresponding to a finite sequence $S$ of elementary row operations, then we call $P:=E_qE_{q-1}…E_2E_1$ the transforming matrix corresponding to $S$.
Theorem: If a finite sequence $S$ of elementary row operations is performed on any matrix $A$, the resulting matrix is $PA$ where $P$ is the transforming matrix corresponding to $S$.
Trivially, I can see that the matrix obtained from $A$ after performing the first elementary row operation (from $S$) is $E_1A$. The matrix obtained after performing the second elementary row is $E_2(E_1A)$, and so on.
However, I wanted a formal proof for the theorem (if there is one).
Best Answer
The matrix obtained after performing the sequence $S$ of row operations on $A$ is $$E_q(E_{q-1}(E_{q-2}...(E_2(E_1A))...).$$ Using transitivity of matrix products,
[as mentioned in @DhanviSreenivasan 's comment: $A(B\cdot C)=(A\cdot B)C$] $$E_q(E_{q-1}(E_{q-2}...(E_2(E_1A))...)=(E_qE_{q-1}...E_2E_1)A$$ If $P$ is the transforming matrix corresponding to $S$, then $P=E_qE_{q-1}...E_2E_1$.
$\therefore$ the resulting matrix after the transformation is $PA$