Let $A$ be a $8 \times 8$ square matrix with $a_{12} = a_{24} = a_{33} = a_{41} = a_{58} = a_{65} = a_{77} = a_{86} = 1$ and all other entries $0$. Then $A^n=I$ for some $n$; find $n$.
I successfully found that $n=3$ by multiplying the matrix $A$ one by one. I want to ask if there is any alternative method that is easier and less tedious.
On
Hint This is a permutation matrix $P_{\sigma}$ for the permutation $\sigma = (124)(586)$. Permutation matrices have the convenient properties that $P_{\sigma \tau} = P_{\sigma} P_{\tau}$ and $P_{\textrm{id}} = I$. Thus, $n$ is the minimal positive $n$ such that $\sigma^n = \textrm{id}$, that is, the order of $\sigma$.
This is a permutation matrix.
We can represent it by
$$(4, 1, 2)(5, 8, 6).$$
which consists of $3$-cycles. Hence $A^n=I$, for $n$ being multiple of $3$.