Number of 3x3 matrices with determinant $1$ and coefficients in $\mathbb{Z}_5$ Let $M=(m_{ij}), m_{ij} \in \mathbb{Z}_5$. 
$det(M) \in \{0,1,2,3,4\}$. There are equal number of matrices with determinants $1,2,3$ and $4$, because determinant is multiplied by $2$ when one of rows is doubled, and is multiplied by $-1$ when two rows are swapped.
The only question is the number of singular matrices. I have written a Python script and found that:
There are 465125 3x3 matrices in GF(5) with det=0
There are 372000 3x3 matrices in GF(5) with det=1
There are 372000 3x3 matrices in GF(5) with det=2
There are 372000 3x3 matrices in GF(5) with det=3
There are 372000 3x3 matrices in GF(5) with det=4

Can you explain why the number of singular matrices is $465125$?
 A: To count the number of singular matrices over a finite field, it suffices to count the number of nonsingular matrices.
To count the number of nonsingular matrices, it suffices to count how may ways you can create a matrix whose rows are linearly independent.
You can pick any nonzero vector $x$ at all to be the first row. ($5^3-1$ choices)
Then you can pick any vector $y$ not in the span of $x$ to be the second vector (there are $5^3-5$) such vectors.
For the third vector $z$, you can pick anything not in the span of $x$ and $y$. Since there are $5^2$ things in the span of $x$ and $y$, you have $5^3-5^2$ choices.
Thus you have $(5^3-1)(5^3-5)(5^3-5^2)=1488000$ nonsingular matrices.
Since there are $5^9$ total matrices, this accounts for $5^9-1488000=465125$ singular matrices.
Now that you've seen this example, I imagine you can adapt the argument to whatever finite field and whatever size matrices that you like :)
(Python's awesome, btw. Just sayin'.)

Added: And as mesel mentions in the comments, the nonzero determinants are split equally among the $1488000$ nonsingular matrices. This is the case because the determinant is a group homomorphism of the group of nonsingular matrices onto the multiplicative group of $F_5$. The kernel of the determinant map is the set of matrices with determinant $1$. The cosets modulo the kernel get mapped to each of the values $\{1,2,3,4\}$, and the cosets are equinumerous with each other, so they all have $\frac{1488000}{4}$ elements.
