Let $p$ be an odd prime number, and consider the set

$ T_{p} \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b\\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of matrices $A \in T_p$ whose determinant is not divisible by $p$.

Recall that for

$ A \;=\; \begin{pmatrix} a & b \\ c & a \end{pmatrix}, $

the determinant is

$ \det(A) \;=\; a^2 \;-\; b\,c. $

Hence $\det(A)$ is not divisible by $p$ precisely when

$ a^2 \;-\; b\,c \;\not\equiv\; 0 \pmod{p}, \quad\text{i.e.,}\quad a^2 \;\neq\; b\,c \pmod{p}. $

Total number of matrices

Since $a,b,c$ each range over $\{0,1,\dots,p-1\}$, there are

$ p^3 $

total possible matrices in $T_p$.

Step 1: Count how many $(a,b,c)$ do satisfy $a^2 \equiv b\,c \pmod{p}$

It will be easier to first count the number of triples $(a,b,c)$ for which

$ a^2 \;\equiv\; b\,c \pmod{p}, $

and then subtract that from $p^3$.

Case A: $a = 0$

Then $a^2 = 0$. We need

$ b\,c \;\equiv\; 0 \pmod{p}. $

Over the field $\mathbf{F}_p$, the product $b\,c\equiv 0$ if and only if at least one of $b$ or $c$ is $0$ (since $p$ is prime).

If $b=0$, then $c$ can be anything in $\{0,\dots,p-1\}$.

This gives $p$ possibilities.

If $c=0$, then $b$ can be anything in $\{0,\dots,p-1\}$.

This also gives $p$ possibilities.

However, the pair $(b=0, c=0)$ is counted in both cases, so we must subtract it out once.

Hence, for $a=0$, the number of $(b,c)$ such that $b\,c \equiv 0$ is

$ p \;+\; p \;-\; 1 \;=\; 2p \;-\; 1. $

Case B: $a \neq 0$

Then $a$ can be any nonzero element in $\{1,2,\dots,p-1\}$; there are $p-1$ such choices. In that scenario, $a^2 \neq 0\pmod{p}$. We want

$ b\,c \;\equiv\; a^2 \pmod{p}. $

Over $\mathbf{F}_p$, a nonzero right-hand side $a^2$ can be written as $b\,c$ in the following way:

If $b=0$, then $b\,c = 0$, which cannot equal $a^2\neq 0$.

No solutions in that subcase.

If $b \neq 0$, then for each nonzero $b$ there is exactly one $c$ satisfying $b\,c \equiv a^2\pmod{p}$, namely $c \equiv a^2 b^{-1}\pmod{p}$.

Since $b$ can be any of the $p-1$ nonzero elements, we get $p-1$ solutions $(b,c)$ for each nonzero $a$.

Thus, for each $a\neq 0$, there are $p-1$ pairs $(b,c)$. Hence, for $p-1$ values of $a$, the total number of solutions in this case is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

Total number of solutions to $a^2 = b\,c$

Summing these two cases:

Case A $(a=0)$: $2p - 1$ solutions.

Case B $(a\neq 0)$: $(p-1)^2$ solutions.

Hence the total count of $(a,b,c)$ satisfying $a^2 \equiv b\,c$ is

$ (2p -1) \;+\; (p-1)^2 \;=\; (2p -1) \;+\; \bigl(p^2 \;-\;2p \;+\;1\bigr) \;=\; p^2. $

(A classic result: there are exactly $p^2$ solutions to $b\,c = a^2$ over $\mathbf{F}_p^3$.)

Step 2: Count how many $(a,b,c)$ satisfy $a^2 \neq b\,c$

Since there are $p^3$ total triples $(a,b,c)$, the number that satisfy

$ a^2 \;\neq\; b\,c \pmod{p} $

is simply

$ p^3 \;-\; p^2. $

This directly corresponds to matrices $A$ whose determinant $a^2 - b\,c$ is $\not\equiv 0\pmod{p}$.

Hence, the number of matrices in $T_p$ with $\det(A)$ not divisible by $p$ is

$ \boxed{p^3 - p^2}. $

The expression $p^3 - p^2$ matches Option D.