Consider matrices $M$ of size $L\times L$ over a finite field $\mathbb{Z}_p$, for simplicity focus on $p$ prime. The size $L$ is even. We want to find the order of a specific class of matrices, namely we want to find the smallest non-zero integer $n$ such that $$M^n=1,$$ where 1 is here the identity matrix of size $L\times L$. The matrix $M$ has a specific block structure, which originates from a specific linear cellular automata. We have $$M=AB,$$ where
$$A=
  \begin{pmatrix}
    1 & 1 & & && & \\
    1 & -1 && & & & \\
   & &    1 & 1 && &\\
   & &  1 & -1 && & \\
   &&&& \ddots & & \\
   &  & & &&1 & 1\\
     &  & & &&1 & -1\\
  \end{pmatrix}$$
and
$$
  B=
   \begin{pmatrix}
 -1 && & & && 1&\\
   &    1 & 1 && && \\
   &  1 & -1 && && \\
  &&& \ddots & & & \\
    & & &&1 & 1&\\
    & & &&1 & -1&\\
    1 && & & && 1\\
  \end{pmatrix}$$

You can see $B$ as $CAC^{-1}$ where $C$ is a cyclic shift over $L$ variables. Separately $A$ and $B$ have simple properties, but their alternating product becomes complicated. 

I did some numerical testing, choosing first $\mathbb{Z}_3$. The interesting thing is (which also motivates me to look deeper into this) that the order $n$ seems to have a complicated behaviour as a function of $L$ and it is not clear to me what is the precise source of this. Sometimes $n$ is very big, seemingly expontentially growing with $L$. For example $n(L=46)=354292=2^2\cdot 23\cdot 3851$. Or $n(L=58)=9565940=2^2\cdot 5\cdot 29\cdot 16493$. But if $n$ is divisible by 6 then we get much lower numbers, $n(60)=120$ for example. I understand that it is natural  to see the prime $p$ somehow reflected in the function $n(L)$, but what I don't understand how the big primes mentioned above can enter the game.