Find a matrix that results in a permutation Apologies for the sort of vague title, but part of my problem is that I'm not quite sure of what my problem actually is asking!
Given a vector $v_n = \begin{pmatrix} 1\\ ... \\n \end{pmatrix} $ 
, for all $\sigma \in S_3$ find a matrix $M_\sigma$ such that $M_\sigma$$v_3 = \begin{pmatrix} \sigma(1) \\ \sigma(2) \\ \sigma(3) \end{pmatrix}$
I think I'm meant to find a matrix that will produce a result that contains permutations of 1, 2, 3 when multiplied by a vector (1, 2, 3), but please correct me if I'm wrong. In addition, how can I go about finding a solution(s)? 
 A: For example, if $\;\sigma 1=2\;,\;\;\sigma2=1\;,\;\;\sigma3=3\;$ , then
$$M_\sigma=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\implies M_\sigma\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}2\\1\\3\end{pmatrix}$$
As you can see, $\;M_\sigma\;$ is just the elementary matrix which interchanges the first and second rows. Try to find other ones for the other elements of $\;S_3\;$
A: Hint:
Let 
$$v = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$$
Consider the identity matrix $I_{3 \times 3}$ and permute its $1$st and $2$nd column. We obtain matrix 
$$I_{(12)}=\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
Try to find $I_{(12)} \cdot v$.
A: Well, $S_3 = \{(1,2,3), (1,3,2), (2, 1,3), (2,3,1), (3, 1, 2), (3,2,1) \}$, so the way the question is stated, you need to find matrices $M_\sigma$ such that $M_\sigma (1,2,3)^T = \sigma$.
$\sigma = (1,2,3)$ is a joke, because this is the Identity matrix. Namely, $I v_3 = v_3$.
Let's take $\sigma = (2,3,1)$. Can we find a $M$ such that $M (1,2,3)^T = (2,3,1)^T$?
You want the second element of (1,2,3) to be on the first place, so the first row becomes (0,1,0). The second row becomes (0,0,1) and the third row (1,0,0). You find :
$$
M_\sigma = M_{(2,3,1)} = \left(
\begin{array}{ccc}
 0 & 1 & 0 \\
 0 & 0 & 1 \\
 1 & 0 & 0 \\
\end{array}
\right)
$$
The other permutations are to be done, but that's left up to you.
