Show that there exists a $3 × 3$ invertible matrix $M$ with entries in $\mathbb{Z}/2\mathbb{Z}$ such that $M^7 = I_3$.
Show that there exists a $3 × 3$ invertible matrix $M$ (which is not the identity matrix) with entries in the field $\mathbb{Z}/2\mathbb{Z}$ such that $M^7 = $Identity matrix.
All I could do was use hit and try method. I was checking different matrices which might satisfy this condition. Of course, it's a bad approach.
Solution 1:
We have the factorization $X^7-1 = (X+1)(X^3+X+1)(X^3+X^2+1)$ over $\mathbb{Z}/2 \mathbb{Z}$. So such matrices are exactly those with characteristic polynomial $X^3+X+1$ or $X^3+X^2+1$—the companion matrix of either polynomial, for example.