For any n you have vector having values for f(n),..,f(n-2),g(n),..,g(n-2),h(n),..,h(n-2) in some order, Then just make a matrix s.t. you get the right vector for n+1 when multiplying that vector from the left with that matrix.

There's one nice trick. Notice 2nd & 3rd equations. Let p(n)=g(n)-h(n):

g(n)=f(n-2)+h(n-1), h(n)=f(n-2)+g(n-1)

Let's subtract: g(n)-h(n)=h(n-1)-g(n-1) => p(n)=-p(n-1) But g(0)=h(0) => p(0)=0. So, p(n)=0 for every n. It wields that g(n)=h(n). So, original system can be rewritten as following:

f(n)=f(n-3)+2g(n-2), g(n)=f(n-2)+g(n-1)

From here it can be easily shown that both of them satisfy the same equation: x(n)=x(n-1)+x(n-3)+x(n-4). Only initial conditions differ. If x=f, then x(0)=x(1)=x(2)=1. If x=g=h, then x(k)=k for k=0,1,2. So, you need only one matrix of order 4 to solve it. By the way, for x=f it's known sequence: OEIS.

Of course, it makes no sense because there is a simple solution as adamant shows =) But I think it's sometimes interesting to understand what's going on behind the scene.