1. Given set

We have

where $p$ is an odd prime.

So total number of matrices in $T_p$ is

since each of $a,b,c$ has $p$ choices.

2. Condition on determinant

For

we have

We want the number of matrices such that $\det(A)$ is not divisible by $p$, i.e.

It is easier to count matrices for which

and subtract from $p^3$.

3. Count solutions of $a^2\equiv bc \pmod p$

We count triples $(a,b,c)$ over $\mathbb F_p$ satisfying

We split into cases.

Case 1: $a=0$

Then the equation becomes

Number of pairs $(b,c)$ with $bc=0$:

• $b=0$: $p$ choices for $c$
• $c=0$: $p$ choices for $b$
• subtract double count $(0,0)$ once

So count is

Thus for $a=0$, number of solutions is

Case 2: $a\neq 0$

Then $a^2\neq 0$, so we need

Hence both $b\neq 0$ and $c\neq 0$.

For a fixed nonzero $a$, and for each nonzero $b$, there is a unique

So number of pairs $(b,c)$ is

Since there are $p-1$ choices for nonzero $a$, total solutions in this case are

4. Total singular matrices modulo $p$

Hence number of triples satisfying $a^2-bc\equiv 0\pmod p$ is

Now

so

Thus the number of matrices with determinant divisible by $p$ is

5. Required count

Therefore the number of matrices with determinant not divisible by $p$ is

6. Compare with options

This matches

which is Option D.

7. Comparison with stored answer

Stored correct answer is D, which agrees with our derived answer.