Is there any way to find the number of matrices with a given determinant in $GL(2,\mathbb{F_{5}})$ I'm trying to find the number of elements of determinant  $3$ in $GL(2,\mathbb{F_{5}})$
The determinant should be $3$ so the product of both the eigenvalues $\lambda_{1}$ and $\lambda_{2}$ will be $3$ , so the choices of pair of eigenvalues are $(1,3),(3,1),(4,2),(2,4)$ because elements should be from $\mathbb{F_{5}}$ .Further i am getting confused how to proceed, because it will be a lengthy and time consuming process to find each and every matrix satisfying those conditions (above), or is there any other suitable method to solve this problem, Please help
Thankyou.
 A: There is a way to find the order of $GL(m,\mathbb F_q)$.
I think the formula was $\prod\limits_{i=0}^{m-1} q^m-q^i$. (because if we select the rows in order there is $q^m-q^i$ options when we select the $i+1$'th row, since it shouldn't be in the span of the previous ones).
Once we have the order of the group we can notice for each non-zero value there is the same number of matrices with that determinant. This is because multiplying the top row by any value multiplies the determinant by the same value, so the equivalence relation between matrices given by "They are the same except for the top row being scalar multiples of each other" splits the matrices into equivalence classes in which there is exactly one determinant for each modulo class
So I think the answer is:
$(\prod\limits_{i=0}^{m-1} q^m-q^i)/(q-1)$
In the case $q=5$ and $m=2$ we get $( (25-1)(25-5))/(5-1) = (24)(20)/4 = 120$
A: @Onir's answer is perfect, but what about @honey kumar's idea? Will it work?
Yes, but it seems quite a lot harder, and it is much easier to make mistakes during it.
For each eigenvalue pair (1,3) and (2,4) we can choose the eigenspaces arbitrarily. That gives $\frac{(5^2-1)(5^2-5)}{(5-1)(5-1)} = 30$. We shouldn't count (1,3) and (3,1) as different since they would give the same matrix using my construction.
But that only gives 60 matrices when the correct answer is 120. :(
That's because the eigenvalues don't have to lie in GF(5), but can be in the larger field GF(25). We also need eigenvalue pairs there. That means we need $x,y \in \operatorname{GF}(25)$ with $xy=3$ AND we need $y=x^5$ so that both $x,y$ are roots of the characteristic polynomial of the matrix. So we need to solve $x^6 = 3$ over GF(25). That gives 3 more, each with 20 matrices, a total of 60 new matrices. (I got 20 using conjugacy classes, I'm not positive how to phrase it using linear algebra.)
