Row Switching Matrix I know that there exists an elementary matrix that switches the rows in another matrix when they are multiplied, but how do you prove that this elementary matrix actually does this job?
 A: If you view a matrix $A$ as an endomorphism $\phi$ of $\mathbb{R}^n$, you have that (called $e_i$ the canonic base) the $i$-th column represents $\phi(e_i)$. It follows that the $i$-th row corresponds to $\psi(e_i)$, where by $\psi$ we denote the endomorphism corresponding to $A^T$. 
Now if you exchange the $i$-th and the $j$-th row, you're basically looking for a permutation matrix (corresponding to the permutation $(i,j)$ ) on $\mathbb{R}^n$ and then using $\psi$. Those permutation matrices are what you are looking for. 
Now, $S_n$ is generated by transpositions and, trivally, exchanging $e_i$ with $e_j$ is accomplished by the matrix $a_{ij}=a_{ji}=1, a_{kk}=1 \; \forall k \neq i,j$ and all other terms $0$. These do what you want on the basis, so they do it on any matrix. 
So for any permutation you might want to do, you have the corresponding matrix obtained by decomposing the permutation in a product of transpositions and multiplying the matrices in the right order. 
A: Let $I(i,j)$ be the identity matrix, except with the $i$th and $jth$ rows reversed. Then by definition of the matrix product, the $(k,l)$th component of the product $I(i,j)A$ is $$(I(i,j)A)_{k,l} = \sum_{m=1}^n (I(i,j))_{k,m}A_{m,l}.$$ If $k \neq i,j$, all the terms in the sum vanish except for the one corresponding to $m = k$, which is $(I(i,j))_{k,k}A_{k,l} = A_{k,l}$.
If $k = i$, say, then again all the terms in the sum vanish except for the one corresponding to $m = j$, which is $(I(i,j))_{k,j}A_{j,l} = A_{j,l}$. Similarly for the case $k = j$.
Thus we see that $(I(i,j)A)$ and $A$ coincide almost everywhere except that the rows $i$ and $j$ are swapped.
