$(A^x) \mathbin\% p = (A^{x \mathbin\% (p - 1)}) % p$ if $p$ is prime. Is this true when $A$ is a matrix?

Question

$(A^x) \mathbin\% p = (A^{x \mathbin\% (p - 1)}) % p$ if $p$ is prime.

Is this property true when $A$ is a matrix?

Suppose $$A=\begin{pmatrix} 1 &0 &1\\ 1 &0 &0\\ 0 &1 &0 \end{pmatrix},$$ $x=1000000007$ and $p=1000000007$

Then $$( A^X ) \mathbin\% p = \begin{pmatrix} 097981108 & 432592815 & 646971665\\ 646971665 & 451009450 & 432592815 \\ 432592815 & 214378850 & 451009450 \end{pmatrix}$$

But $(A^1)\mathbin\%p$ [ as,$ ( x \mathbin\% (p-1) )=1 $ ] is the same matrix.

Answer 1

As Lord Farin points out, this is certainly false. Consider for example any prime $p$ and the matrix $$A=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$$ which satisfies $A^x=0$ for all $x>1$.