On the number of $2 \times 2$ matrices over ${\Bbb Z}_3$ with unit determinant

Question

Determine the number of matrices in ${\Bbb Z}_3^{2 \times 2}$ with determinant $1$.

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I know that the elements in ${\Bbb Z}_3$ are $\{0,1,2\}$ now possibly determinant can be $1$ in this case

$$ \begin {bmatrix} 1&2\\ 0&1\\ \end{bmatrix} $$

and there can be many more but how to find the exact number of such matrices? I know that the answer is $24$.

Answer 1

Let your matrix be $$\pmatrix{a & b\cr c & d\cr}$$ If $b c = 0$ you want $ad = 1$. How many possibilities for that?

If $b c = 2$ you want $ad = 0$. How many possibilities for that?

If $b c = 1$ you want $ad = 2$. How many possibilities for that?

Answer 2

First we count the non-singular $2\times 2$ matrices. The first row can be any of the $8$ non-zero vectors. Then the second row can be anything but a multiple of the first row. There are $3$ such multiples. Thus there are $(8)(6)$ non-singular $2\times 2$ matrices.

Mutiplying a row by $2$ multiplies the determinant by $2$, giving a bijection between matrices with determinant $1$ and those with determinant $2$. So there are $24$ with determinant $1$.

Remark: The idea generalizes to larger matrices, and other finite fields.