Matrices with Trace Not Divisible and Determinant Divisible by Prime

To find the number of matrices $A \in T_p$ such that the trace of $A$ is not divisible by $p$ but $\det(A)$ is divisible by $p$, we analyze the given conditions step-by-step.

The set $T_p$ consists of matrices of the form:

$$A = \begin{pmatrix} a & b \\ c & a \end{pmatrix}, \quad \text{where } a, b, c \in \{0, 1, 2, \dots, p-1\}$$

1. Condition 1: $\text{trace}(A)$ is not divisible by $p$
   The trace of $A$ is given by:
   $\text{trace}(A) = a + a = 2a$
   Since $p$ is an odd prime, $\gcd(2, p) = 1$. Thus, $\text{trace}(A)$ is divisible by $p$ if and only if $a$ is divisible by $p$.
   Since $a \in \{0, 1, 2, \dots, p-1\}$, the condition that $\text{trace}(A)$ is not divisible by $p$ is equivalent to:
   $a \neq 0$
   Therefore, there are $p - 1$ choices for $a$, namely $a \in \{1, 2, \dots, p-1\}$.

2. Condition 2: $\det(A)$ is divisible by $p$
   The determinant of $A$ is:
   $\det(A) = a^2 - bc$
   The condition that $\det(A)$ is divisible by $p$ translates to the modular congruence:
   $a^2 - bc \equiv 0 \pmod p \implies bc \equiv a^2 \pmod p$

3. Counting the number of pairs $(b, c)$ for a fixed $a \neq 0$:
   Since $a \in \{1, 2, \dots, p-1\}$, we have $a \not\equiv 0 \pmod p$, which implies $a^2 \not\equiv 0 \pmod p$.
   Therefore, $bc \equiv a^2 \pmod p$ means:

   • $b$ cannot be $0$, because if $b = 0$, then $bc = 0 \not\equiv a^2 \pmod p$. Thus, $b \in \{1, 2, \dots, p-1\}$.
   • For each of the $p - 1$ possible choices of $b \in \{1, 2, \dots, p-1\}$, the element $b$ is invertible modulo the prime $p$. Its modular inverse $b^{-1} \pmod p$ exists uniquely in $\{1, 2, \dots, p-1\}$.
   • Hence, for every fixed $b$, there exists a unique $c \in \{1, 2, \dots, p-1\}$ given by:
     $c \equiv a^2 \cdot b^{-1} \pmod p$
   • Thus, for any fixed $a \in \{1, 2, \dots, p-1\}$, there are exactly $p - 1$ pairs of $(b, c)$ that satisfy the congruence.

4. Total Number of Matrices:
   Since there are $p - 1$ choices for $a$ and for each choice of $a$, there are $p - 1$ pairs of $(b, c)$, the total number of such matrices $A$ is:
   $\text{Total} = (p - 1) \times (p - 1) = (p - 1)^2$

Hence, the correct option is C.