Homomorphism between symmetric group and general linear group of order n. I am having trouble proving the following:

Show that $f: S_n \to GL_n(\mathbb{R}),\;\: f(x)=A_x$ is a homomorphism where $A_x$ is the permutation matrix associated with $x$. 

$S_n$ is the symmetric group and $GL_n(\mathbb{R})$ is the general linear group of order $n$.
 A: I can explicitly show you how to do this for $S_3$:
$e \to I$, the 3x3 identity matrix.
$(1\ 2) \to \begin{bmatrix}0&1&0\\1&0&0\\0&0&1 \end{bmatrix}$
$(1\ 3) \to \begin{bmatrix}0&0&1\\0&1&0\\1&0&0 \end{bmatrix}$
$(2\ 3) \to \begin{bmatrix}1&0&0\\0&0&1\\0&1&0 \end{bmatrix}$
$(1\ 2\ 3) \to \begin{bmatrix}0&0&1\\1&0&0\\0&1&0 \end{bmatrix}$
$(1\ 3\ 2) \to \begin{bmatrix}0&1&0\\0&0&1\\1&0&0 \end{bmatrix}$
This map assumes you multiply permutations "composition-wise" (right-to left). If you do it the other way, use the transposes.
A: For an $n\times n$ matrix $A$ and a permutation $\sigma$, let $\sigma A$
denote the matrix obtained by using $\sigma$ to permute the rows of $A$. Then
$A_{\sigma}=\sigma I$ and we have to show that $A_{\sigma}A_{\tau}%
=A_{\sigma\tau}$.
Explicitly, if $A=(a_{ij})$, then $\sigma A=(a_{ij}^{\prime})$ with
$a_{ij}^{\prime}=a_{\sigma^{-1}(i)j}$. Use this to show that $(\sigma
A)B=\sigma(AB)$. Now
$A_{\sigma}A_{\tau}=(\sigma I)(\tau I)=\sigma(I\cdot\tau I)=\sigma(\tau
I)=A_{\sigma\tau}.$
