Wedderburn decomposition How can I get the Wedderburn decomposition of the module $KG$, if I have the irreducible represenations of the group $G$? I am confused by a post here I found:
Why is the Wedderburn formula in this case wrong?
The last summand in the Wedderburn decomposition is $Mat(2,F_9)$, but the associated irreducible representation is 
$$< \begin{pmatrix} \ & \ & \ & 1  \\ \ & \ & 1 & \ \\  \ & 1 & \ & \  \\ 1 & \ & \ & \ \end{pmatrix} , \begin{pmatrix} \ & 1 & \ & \  \\ \ & \ & 1 & \ \\  \ & \ & \ & 1  \\ -1 & -1 & -1 & -1 \end{pmatrix} >.  $$
I.E. this are matrices in $Mat(4,F_{3})$ (this is linked in the post). Why do I write than $Mat(2,F_9)$ as a summand?
 A: Since this question's context is mostly a bunch of things I've written, let me mention that everything the OP says is true and fine. The algebra generated by those two matrices is isomorphic to $M_2(\mathbb{F}_9)$.  The center of this algebra is its centralizer, and is isomorphic to $\mathbb{F}_9$.  You then find a set of matrix units inside the ring.  The specific elements do not appear to be educational in this particular case.  Writing $M_2(\mathbb{F}_9) \leq M_4(\mathbb{F}_3)$ is called a "blow up" in this part of algebra, and is simply realizing the isomorphism of natural modules $\mathbb{F}_9^2 \cong \mathbb{F}_3^4$.
Here is a little GAP code to get you started. Center and Centralizer are useful. I'm not sure of the simplest command to get the matrix units.
gap> x:=PermutationMat((1,4)(2,3),4,GF(3));
[ [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3)^0 ], [ 0*Z(3), 0*Z(3), Z(3)^0, 0*Z(3) ], 
  [ 0*Z(3), Z(3)^0, 0*Z(3), 0*Z(3) ], [ Z(3)^0, 0*Z(3), 0*Z(3), 0*Z(3) ] ]
gap> y:=CompanionMat((X(GF(3))^5-1)/(X(GF(3))-1));
[ [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3) ], [ Z(3)^0, 0*Z(3), 0*Z(3), Z(3) ], 
  [ 0*Z(3), Z(3)^0, 0*Z(3), Z(3) ], [ 0*Z(3), 0*Z(3), Z(3)^0, Z(3) ] ]
gap> a:=AlgebraWithOne(GF(3),[x,y]);
<algebra-with-one over GF(3), with 2 generators>
gap> Size(a);
6561
gap> Size(GF(9)^[2,2]);
6561
gap> Size(GF(3)^[4,4]);
43046721

