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.

If you're curious to know all matrices having order $8$, that can be done by brute force. In Python:
# Returns True if m is the 2x2 identity matrix.
def one(m):
  return np.array_equal(m, np.identity(2))

# Returns the square of m, reduced modulo 3.
def square(m):
  return np.vectorize(lambda n: n % 3)(np.dot(m, m))

# Iterate over all 81 2x2 matrices with values in F_3.
for (a,b,c,d) in itertools.product(range(3), repeat=4):
  if (a or b or c or d) == 2:
    # Choose the +/- representative that starts with a 1.
    continue
  m = np.array([[a,b],[c,d]])
  m_2 = square(m)
  m_4 = square(m_2)
  m_8 = square(m_4)
  if one(m_8) and not one(m_4):
    print '\\pm\\pmatrix{%d&%d\\\\%d&%d},\\;' % (a,b,c,d)

This outputs $12$ solutions:
$
\pm\pmatrix{0&1\\1&1},\;
\pm\pmatrix{0&1\\1&2},\;
\pm\pmatrix{1&1\\1&0},\;
\pm\pmatrix{1&1\\2&1},\;
\pm\pmatrix{1&2\\1&1},\;
\pm\pmatrix{1&2\\2&0}
$
