Elementary operations on a matrix and multiplication by elementary matrices If we have a matrices
$$A=\begin{bmatrix} a & b\\ c & d\end{bmatrix} 
\hskip5mm \mbox{ and } \hskip5mm 
e_{12}(\lambda)= 
\begin{bmatrix} 
1 & \lambda \\
0 & 1
\end{bmatrix}
$$
then by doing product
$$
Ae_{12}(\lambda)  = 
\begin{bmatrix} 
a & a\lambda + b\\
c & c\lambda + d 
\end{bmatrix} \hskip5mm \mbox{ and } \hskip5mm 
e_{12}(\lambda)A = 
\begin{bmatrix} 
a +c\lambda & b+d\lambda \\
c &  d 
\end{bmatrix}
$$
we can interpret that right multiplication by $e_{12}$ to $A$ gives a column-operation: add $\lambda$-times first column to the second column.
In similar way, left multiplication by $e_{12}(\lambda)$ to $A$ gives row-operation on $A$.
Question: Is there any conceptual (not computational, if any) way to see that elementary row and column operations on a matrix can be expressed as multiplication by elementary matrices on left or right, accordingly?
 A: Let $A,B$ be two matrices of order $n$.
We can describe $B$ as $B=(b_1,\ldots,b_n)$, where $b_i$ is its column $i$.
Notice that $AB=(Ab_1,\ldots,Ab_n)$.
Now apply some column-operation on $AB$.
For example, let's say that its column $i$ is multiplied by $\lambda$.
So we obtain $(Ab_1,\ldots,\lambda Ab_i,\ldots,Ab_n)$.
Notice that $(Ab_1,\ldots,\lambda Ab_i,\ldots,Ab_n)=A(b_1,\ldots,\lambda b_i,\ldots,b_n)$.
So in order to apply some column-operation on $AB$, we can first apply it on $B$ and then multiply the resulting matrix with $A$.
Can you see what happens if $B=Id$?
The same reasoning can be used for row-opperations.
A: Here conceptual and computational ideas go hand in hand. We can see this by looking at the multiplication of $A$ with $e_{12}(\lambda)$ in some detail.
We have
\begin{align*}
\begin{bmatrix}1&\lambda\\
0&1
\end{bmatrix}
&=\begin{bmatrix}1&0\\
0&1
\end{bmatrix}
+
\begin{bmatrix}0&\lambda\\
0&0
\end{bmatrix}
=I+\lambda\begin{bmatrix}0&\color{blue}{1}\\
0&0
\end{bmatrix}
\end{align*}
It is the position $_{12}$ of the blue marked $1$ which determines selected row resp. column of $A$.
We obtain
\begin{align*}
Ae_{12}(\lambda)&=A\left(I+\lambda
\begin{bmatrix}0&\color{blue}{1}\\
0&0
\end{bmatrix}\right)
=A+\lambda\begin{bmatrix}
0&\color{blue}{a}\\
0&\color{blue}{c}
\end{bmatrix}\\
e_{12}(\lambda)A&=\left(I+\lambda
\begin{bmatrix}0&\color{blue}{1}\\
0&0
\end{bmatrix}\right)A
=A+\lambda\begin{bmatrix}
\color{blue}{c}&\color{blue}{d}\\
0&0
\end{bmatrix}
\end{align*}
