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 automaton. 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 exponentially 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 $L$ 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 is how the big primes mentioned above can enter the game.
Update: Further numerical experiments showed that the smallest orbit happens when $L=2p^m$, in which case $n=2L$, we have simple linear growth. It gets complicated and quickly growing for most other $L$.
The origin of the problem is the following cellular automaton. Consider $L$ variables $s_j\in \mathbb{Z}_p$, with $j=1,2,\dots,L$. We define a dynamics on the model, such that we perform two operations cyclically. In each operation we group two neighbouring variables into a pair, and we alternate the ways how we make the pairing. For each pairing we perform the update
$$(s_j,s_{j+1})\to (s_j+s_{j+1},s_j-s_{j+1})$$
What changes is that at every second update $j$ is even or odd. If you represent this linear operation with matrices, you get $A$ and $B$ and we want to iterate the product $M=AB$.
Best Answer
Big primes enter the picture as follows: Compute the characteristic polynomial $P_L$ of your square matrix $M$ of size $L$. Factor $P_L$ over your working prime $p$. This gives you the eigenvalues of the involved Jordan-blocs over $\mathbb F_p$. These eigenvalues are elements of field extensions of degree at most $L$ of your groundfield $\mathbb F_p$. You can thus get large primes dividing the order of $M$ if they are divisors of $p^k-1$ for $k$ the degree of (at least) one irreducible polynomial (over $\mathbb F_p$) dividing $P_L$.
A fast way for computing the order of $M$ is thus to compute the characteristic polynomial $P_L$ of $M$, factor it over $\mathbb F_p$ and check then if prime-divisors of $p^k-1$ (for $k$ the degree of an involved irreducible polynomial) divide the order.
The order of $M$ over $\mathbb F_p$ is necessarily a divisor of $U=p^{L-1}\prod_{k=1}^{L}(p^k-1)$. One can of course remove a factor $p^k-1$ from this product if $P_L$ mod $p$ has no irreducible divisor modulo $p$ of degree $k$. One can further reduce the size of $U$ by taking greatest common divisors of all involved factors. The exponent $L-1$ of $p$ in $U$ can of course be replaced by $\mu-1$ where $\mu$ is largest occuring multiplicity among the irreducible divisors of $P_L \pmod p$. (There seem indeed to be often multiplicities in $P_L$: This leads to possibly non-trivial Jordan-blocks in $M$ which contribute a factor $p^d$ to the order of $M$ where $d+1$ is the dimension of the largest involved Jordan bloc.)
The computational bottleneck of this approach is the factorization of $p^k-1$. The rest is computationally easy using fast exponentiation. Concretetely, You have to check if $M^{U/q}$ is still the identity for $q$ any prime divisor of $U$. If this is the case, replace $U$ by $U/q$ (and recheck if you can remove one more power of $q$ from $U$). Going through all prime divisors of $U$ gives you at the end the exact order of $M$ over $\mathbb F_p$.
Experimental observation The characteristic polynomial $P_L$ of the square matrix $M=AB$ of even size $L$ seems to be given by $$P_L=(x^2+4)Q^2_{L/2}$$ where $Q_1=1,\quad Q_2=x+2$ and $Q_n=(x+2)Q_{n-1}-xQ_{n-2}$ for $n\geq 3$. (The polynomials $Q_n$ are 'almost' orthogonal polynomials and seem to satisfy the divisibility property $P_a\vert P_b$ if $a\vert b$.) This formula holds for $L\leq 60$ even.