Justification for elementary row operations as a product I am currently learning linear algebra in the first year of undergraduate student. I need to know why the following is true:
Let A be a matrix. Let $f$ be a function that transforms A by some row operation. Then $f($A$)=f($I$)·$A, where I is the identity 
matrix.
So basically, if I perform a row operation on A, why is true that applying the same row operation in the identity matrix, then multiplying A by the resulting matrix by the left leaves the same result?
I've searched on wikipedia and I saw this page but there is no proof for it; neither my textbook has it. 
My teacher doesn't want to explain me this because he thinks it's tedious to prove but I find this very important in order to make sense of further results.

I would be very grateful if someone explains this to me in an intuitive way but keeping the rigor if necessary.
 A: Let $f(I)$ be the $n\times n$ matrix obtained from the $n\times n$ identity matrix $I$ by swapping the $k$-th and $l$-th row. Let $A=(a_{ij})$ be a general $n\times n$ matrix with columns $c_1,\dots,c_n$. So $c_j=(a_{1j}~a_{2j}~\cdots~a_{nj})^\text{T}$.
If $r_i$ is the $i$-th row of $f(I)$, then the $i$-th row of $f(I)A$ is given by
$$\begin{pmatrix}r_ic_1 & r_i c_2 & \cdots & r_i c_n\end{pmatrix}.$$
For $i\neq k,l$   the row $r_i$ of $f(I)$ has 1 on position $i$, so the only summand of $r_ic_j$ that is not $0$ is the one corresponding to the multiplication of $i$-th positions in $r_i$ and $c_j$. The $i$-th position in $c_j$ is $a_{ij}$ so $r_ic_j=1\cdot a_{i,j}$. Therefore, the $i$-th row of $f(I)A$ is
$$\begin{pmatrix}a_{i1} & a_{i2} & \cdots & a_{in}\end{pmatrix},$$
which is the same as the $i$-th row of $A$.
The row $r_k$ has $1$ at position $l$ so the only summand of $r_kc_j$ that is not $0$ is  the one corresponding to the multiplication of $l$-th positions in $r_k$ and $c_j$. The column $c_j$ has $a_{lj}$ as its $l$-th position so $r_kc_j=a_{lj}$ therefore the $k$-th row of $f(I)A$ is
$$\begin{pmatrix}a_{l1} & a_{l2} & \cdots & a_{ln}\end{pmatrix},$$
which is the $l$-th row of $A$.
Similarly, the $l$-th row of $f(I)A$ is
$$\begin{pmatrix}a_{k1} & a_{k2} & \cdots & a_{kn}\end{pmatrix},$$
which is the $k$-th row of $A$.
A: Well, with your notation, this is just saying that $EA=E(IA)=(EI)A$ (since $IA=A$).
