<s>[INST] 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.
 [/INST] Using a similar strategy to Travis, we can actually answer this question quite nicely: 
$$A \text{ has order 8 if and only if }\det(A) = 2 \text{ and } \text{tr}(A) \neq 0.$$ We can do this by looking at the characteristic polynomial $p(x)$ of $A$, with which we can utilize the fact that $p(A) = 0$ as shown below.
Since $A$ is a $2\times 2$ matrix, this takes the simple form $p(x) = x^2 - tx + d$ where $t$ is the trace of $A$ and $d$ the determinant (which cannot be 0 since $A$ is invertible). If the trace is 0, then $A^2 = dI$ and so has order $2$ if $d = 1$ and 4 if $d = 2$. Thus if we are looking for an invertible matrix $A$ with order 8, we know that $d$ and $t$ must be nonzero.
We can now calculate powers of $A$ quite readily through successive squaring, use two key observations to simplify the calculations: 1)in $\mathbb{F}_3$, 2 is the same as $-1$ and so we can replace any appearance of 2 with a negative sign and 2) since $t$ and $d$ are both either $1$ or $2 = -1$, it is always true that $t^2 = d^2 = 1$. The calculations in $\mathbb{F}_3$ are as follows, using the fact that $p(A) = 0$: 
$$A^2 = tA-dI$$
$$ A^4 = (tA-dI)^2 = A^2+tdA + I = t(1+d)A + (1-d)I,$$
$$ A^8 = (1+d)^2A^2+(1-d)^2I =  -(d+1)A^2 + (d-1)I$$
$$\implies A^8 = [d(d+1)+(d-1)]I - t(d+1)A = -dI - t(d+1)A.$$
We can therefore conclude that the order of $A$ is 8 if and only if $-d = 1$ and $t(d+1) = 0$, which since $t \neq 0$ are both equivalent to $d = 2$. Therefore, an invertible matrix $A \in GL(2,\mathbb{F}_3)$ has order 8 iff $\det(A) = 2$ and tr$(A) \neq 0$ (which we showed is necessary before). You'll notice that this agrees with the concrete examples given by Travis and Chris Culter.
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