Prove that elementary matrices perform row operations How to prove that elementary matrices actually perform their intended row operations: multiplying by a constant, adding a multiple of one row to another, and switching two rows? 
I've seen examples of their use, but I haven't seen a proof for an $n$ by $n$ matrix.  
 A: The key is to understand what matrix matrix multiplication really does. Consider the problem of computing the product $C = AB$, where all matrices are square of dimension $n$ for the sake of simplicity. By definition we have
\begin{equation}
c_{i,j} = \sum_{k=1}^n a_{ik}b_{k,j}
\end{equation}
for each component $c_{i,j}$. If we zoom out and consider the $i$ row of $C$ using notation adapted from MATLAB we have
\begin{equation}
c_{i,1:n} = \sum_{k=1}^n a_{ik}b_{k,1:n}.
\end{equation}
This shows that the $i$th row of $C$ is obtained by forming linear combinations of all $n$ rows of $B$ using the coefficient along the $i$th row of $A$ as weights. In particular, if $a_{ik} = 0$, then the $k$ row of $B$ is not involved in the computation of the $i$th row of $C$.
The case of the elementary matrices is now easy to understand as the vast majority of the entries are either zeros or ones.
A: Let's go through an example of proving that one kind of elementary matrix does what it's supposed to.
In particular, let $E$ be the elementary matrix that adds $\beta$ times the first row of an $n \times k$ matrix to the second row.  We will know that $E$ does what it's supposed to do if, for an $n \times k$ matrix $A$, we have
$$
(EA)[i,j] = \begin{cases}
a_{ij} & i \neq 2\\
\beta a_{1j} + a_{2j} & i =2 
\end{cases} \tag{*}
$$
So, now I'm going to say what my matrix $E$ is, then show that $E$ does what it's supposed to, which is to say that it satisfies $(*)$.
My $E$ will be given by
$$
e_{ij} = 
\begin{cases}
1 & i=j\\
\beta & i=2, \quad j=1\\
0 & \text{otherwise}
\end{cases}
$$
Pro-tip: to find $E$ for a given row operation, just apply the row-operation to the identity matrix and use the matrix that you get.
Now, let's see what $(EA)[i,j]$ is, using the definition of matrix multiplication: first, the case that $i \neq 2$.  Note that $e_{ik} \neq 0$ only if $i = k$.  So, we have
$$
(EA)[i,j] = 
\sum_{k=1}^n e_{ik}a_{kj} = e_{ii}a_{ij} = a_{ij}
$$
Now, if $i = 2$, we have $e_{ik} \neq 0$ except at $i = 1$ and at $i=2$.  So, we find
$$
(EA)[2,j] = \sum_{k=1}^n e_{ik}a_{kj} = e_{21}a_{1j} + e_{22}a_{2j} = \beta a_{aj} + a_{2j}
$$
So, we see that the matrix $E$ indeed does exactly what it was supposed to do.
