A matrix of order 8 over $\mathbb{F}_3$ What is an example of an invertible matrix of size 2x2 with coefficients in $\mathbb{F}_3$ that has exact order 8? 
I have found by computation that the condition that the 8th power of a matrix 
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is the identity is
$$
b c (a + d)^2 (a^2 + 2 b c + d^2)^2 + ((a^2 + b c)^2 + 
   b c (a + d)^2)^2=1, \qquad b (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 + 
   4 a b c d + 4 b c d^2 + d^4)=0, \qquad c (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 + 
   4 a b c d + 4 b c d^2 + d^4)=0, \qquad b c (a + d)^2 (a^2 + 2 b c + 
    d^2)^2 + (b c (a + d)^2 + (b c + d^2)^2)^2=1
$$
and the condition for invertibility is $ad\neq bc$. If the 4th power is not the identity, then no power that is not a multiple of 8 is not the identity (because we could cancel out to either get that the first power is the identity or that the second power is the identity, both lead to contradiction). That is another cumbersome condition to write out.
I hope somebody can suggest a nicer way.

We can use the same approach but reduce drastically the complexity of the system in the entries $a, b, c, d$ if we instead look for a square root of some matrix with order $4$.
The matrix $A = \pmatrix{0&-1\\1&0}$ satisfies $A^2 = -I$ and so has order $4$ over any field of characteristic not $2$. In particular, if we can find a matrix $B$ such that $B^2 = A$, then $B$ will have order $8$.

Writing $B = \pmatrix{a&b\\c&d}$, the condition is equivalent to the system \begin{align} a^2 + bc &= 0\\ b(a + d) &= -1\\ c(a + d) &= 1\\ d^2 + bc &= 0 \end{align} The second equation implies that $b = \pm 1$, and by negating the matrix (which preserves the property that $B^2 = I$) we may assume $b = 1$, and substituting gives $a + d = -1$ and $c = -1$. The first and fourth equations give that $a, d = \pm 1$, and then the third equation gives that $$a = d = 1 ,$$ yielding the solution $$\pmatrix{1&1\\-1&1} .$$

