believe it or not, the way this comes about is starting with a three-dimensional representation:

(the other 4 matrices can be obtained in the usual way from these generators).

one can then ask: what subspace in the standard basis {e_{1},e_{2},e_{3}} is invariant under (all six of) these mappings?

one answer is the hyperplane {x_{1}e_{1}+ x_{2}e_{2}+ x_{3}e_{3}: x_{1}+x_{2}+x_{3}= 0}.

note that the defining equation of the hyperplane is the (elementary) symmetric polynomial of degree 1 in 3 variables.

since we have an invariant subspace, we might consider the induced action of S_{3}on this subspace.

to come up with matrices we, of course, need to pick a BASIS for this subspace. there are various ways to do this (each one yielding different matrices).

well e_{1}- e_{2}= (1,-1,0), and 1 + (-1) + 0 = 1, so this is one possible basis vector. another possible basis vector is e_{1}- e_{3}= (1,0,-1).

since a hyperplane in a n-dimensional space has dimension n-1, these are all we need. now let's look at which 2x2 matrices we get in the basis B = {e_{1}-e_{2},e_{1}-e_{3}}.

so let's look at what the 3x3 matrices do to e_{1}- e_{2}, and e_{1}- e_{3}.

the 3x3 matrix for (1 2) sends e_{1}- e_{2}to e_{2}- e_{1}. so it sends the vector [1,0]_{B}to [-1,0]_{B}.

the 3x3 matrix for (1 2) sends e_{1}- e_{3}to e_{2}- e_{3}= -(e_{1}- e_{2}) + e_{1}- e_{3},

so it sends the vector [0,1]_{B}to [-1,1]_{B}. thus in the basis B, the induced representation of S_{3}on our hyperplane gives for (1 2) the matrix:

.

now let's do the same thing for (1 2 3):

e_{1}- e_{2}gets sent to e_{2}- e_{3}= -(e_{1}- e_{2}) + e_{1}- e_{3}, so (1 2 3) sends [1,0]_{B}to [-1,1]_{B}

e_{1}- e_{3}gets sent to e_{2}- e_{1}= -(e_{1}- e_{2}) so (1 2 3) sends [0,1]_{B}to [-1,0]_{B},

which gives us for (1 2 3) in the basis B the matrix:

.