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.

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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.

I don’t know how enlightening it is to literally just have an example of such a matrix to “understanding” the structure of $GL_3(\Bbb F_2)$, but for what it’s worth,

$$A=\left(\begin{matrix}0&0&1\\0&1&1\\1&1&0\end{matrix}\right)$$

is such a matrix, found by enumerating all the matrices (there are only $512$ so it’s not so bad).