[Robert wrote](https://forum.azimuthproject.org/discussion/comment/18891/#Comment_18891):
> I guess we'd be looking for the sum of the numbers on the resulting main diagonal: the trace of the matrix.

The trace would give you the number of paths of length \$$n\$$ either of type \$$x \to x\$$ or of type \$$y \to y\$$. For instance, the square of that matrix gives:

$\left( \begin{array}{cc} 2 & 1 \\\ 1 & 1 \end{array} \right) .$

The trace of this matrix is 3, but we know there are only two paths of length 2 from x to itself. Do you see what's happening? (It has to do with the fact that the original matrix is, as you observed, the _adjacency matrix_ for this graph.)

Also, notice that the adjacency matrix doesn't have to be binary-valued here. If we have two self-loops on a vertex, we'd need to put a 2 down for that vertex.

(I love the matrix approach to this problem! It's a wonderful example of one of [Pólya's principles](https://en.wikipedia.org/wiki/How_to_Solve_It): sometimes, in order to prove a thing, you should prove a stronger thing instead.)