Need help understanding matrix representations of the symmetric group $S_3$. I have the following map for a representations of $S_3$:
$$e \mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 2) \mapsto 
\begin{pmatrix}
0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 3)\mapsto 
\begin{pmatrix}
0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0
\end{pmatrix}$$
$$(2\; 3)\mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0
\end{pmatrix}, \quad (1\; 2\; 3) \mapsto 
\begin{pmatrix}
0& 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{pmatrix}, \quad (1\; 3\; 2)\mapsto 
\begin{pmatrix}
0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0
\end{pmatrix}$$
We can check that any $\sigma \in S_3$ and its image under the map represents the same permutation. For example, consider multiplying the matrix associated with $(2\; 3)$ with the column vector $[a\; b\; c]$:
$$\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0
\end{pmatrix} \begin{pmatrix}
a \\ b \\ c
\end{pmatrix} = \begin{pmatrix}
a\\c\\b
\end{pmatrix}$$
The second and third element in the column vector are interchanged, with the first element remaining fixed. This is the kind of behavior I expected any representation of $S_3$ will be exhibit. However, when I consider the "standard representation" of $S_3$, given as:
$$e \mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 2) \mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 3)\mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0
\end{pmatrix}$$
$$(2\; 3)\mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 &  \\ 0 & 1 & -1
\end{pmatrix}, \quad (1\; 2\; 3) \mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1
\end{pmatrix}, \quad (1\; 3\; 2)\mapsto 
\begin{pmatrix}
1 & 0 & 0 \\ 0 & -1 & 1 \\ 0& -1 & 0
\end{pmatrix}$$
I don't know how to interpret the result I get from looking at it the way I was looking above, using matrix multiplication. For instance,
$$
\begin{pmatrix}
1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
a\\b\\c\end{pmatrix}= \begin{pmatrix}
a\\-b+c\\c
\end{pmatrix},$$
from which I have no idea what to infer. 
I realize I'm not explaining myself very well, but I hope someone can maybe take a stab at answering my question anyway. Thank you very much for your help.

Your two representations are equivalent, the permutation representation, they just use different bases to represent linear operators as different matrices. The first uses the standard basis. 
The permutation representation of $S_3$ is reducible; the span of $(1,1,1)$ is an invariant subspace, as is its complement comprised of $(x,y,z)$ satisfying $x+y+z=0$. Since decomposition into irreducible representations is the primary focus of representation theory, the 2D subspace is what's actually called the "standard representation". In general, the permutation representation of $S_n$ is a direct sum of a 1D trivial subrepresentation and the standard representation of dimension $n-1$, just like for $S_3$.
In the second set of matrices you present, the basis $\{(1,1,1),(1,-1,0),(0,1,-1)\}$ is used instead. To calculate the $2\times2$ part of the matrix for say $(12)$, write out
$$ (12)\left(\color{red}{a}\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\color{blue}{b}\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix}\right)=a\begin{bmatrix}-1\\ \phantom{+}1 \\ \phantom{+}0\end{bmatrix}+b\begin{bmatrix}\phantom{+}1\\ \phantom{+}0\\-1\end{bmatrix} $$
$$ = -a\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+b\left(\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix}\right)=\color{green}{(-a+b)}\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\color{purple}{b}\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix} $$
which matches
$$ \begin{bmatrix} -1 & 1 \\ \phantom{+}0 & 1 \end{bmatrix} \begin{bmatrix} \color{red}{a} \\ \color{blue}{b} \end{bmatrix} = \begin{bmatrix} \color{green}{-a+b} \\ \phantom{+}\color{purple}{b} \end{bmatrix}. $$
(Apologies to the color-blind.)
These $2\times2$ matrices also represent the possible permutations of $\{0,1,\infty\}$ in the Riemann sphere using Mobius transformations.
