Number of Invertible Matrices over Finite Field

From the passage associated with this problem, the set $T_p$ is defined for an odd prime $p$ as:

$$T_p = \left\{ A = \begin{pmatrix} a & b \\ c & a \end{pmatrix} : a, b, c \in \{0, 1, 2, \dots, p-1\} \right\}$$

1. Total Number of Matrices in $T_p$

Each of the elements $a, b, c$ can independently take any of the $p$ values in $\{0, 1, 2, \dots, p-1\}$. Therefore, the total number of matrices in $T_p$ is:
$|T_p| = p \times p \times p = p^3$

2. Condition for $\det(A)$ to be Divisible by $p$

The determinant of matrix $A$ is given by:
$\det(A) = a^2 - bc$
The condition that $\det(A)$ is divisible by $p$ is equivalent to:
$a^2 - bc \equiv 0 \pmod p \implies a^2 \equiv bc \pmod p$

We count the number of triples $(a, b, c)$ satisfying this congruence:

• Case 1: $a = 0$
  Here, $a^2 = 0$, so we must have:
  $bc \equiv 0 \pmod p$
  Since $p$ is a prime number, this means either $b \equiv 0 \pmod p$ or $c \equiv 0 \pmod p$:

  • If $b = 0$, $c$ can be any of the $p$ elements in $\{0, 1, \dots, p-1\}$ ($p$ solutions).
  • If $b \neq 0$ ($p - 1$ choices), then $c$ must be $0$ ($1$ choice).

  Thus, the number of pairs $(b, c)$ when $a = 0$ is:
  $p + (p - 1) = 2p - 1$

• Case 2: $a \neq 0$
  There are $p - 1$ possible choices for $a \in \{1, 2, \dots, p-1\}$.
  For each fixed value of $a$, $a^2 \not\equiv 0 \pmod p$.
  We need:
  $bc \equiv a^2 \pmod p$
  Since $a^2 \not\equiv 0 \pmod p$, neither $b$ nor $c$ can be $0$.
  For any non-zero choice of $b \in \{1, 2, \dots, p-1\}$ ($p - 1$ choices), there exists a unique multiplicative inverse $b^{-1} \pmod p$. Hence, $c$ is uniquely determined modulo $p$ as:
  $c \equiv a^2 b^{-1} \pmod p$
  Therefore, for each of the $p - 1$ choices of $a$, there are $p - 1$ choices for $b$, and $c$ is uniquely determined.
  The number of triples in this case is:
  $(p - 1) \times (p - 1) = (p - 1)^2$

3. Total Number of Matrices with $\det(A)$ Divisible by $p$

Adding the results from both cases:
$N_{\text{divisible}} = (2p - 1) + (p - 1)^2 = 2p - 1 + p^2 - 2p + 1 = p^2$

4. Number of Matrices with $\det(A)$ Not Divisible by $p$

The number of matrices in $T_p$ such that $\det(A)$ is not divisible by $p$ is:
$N = |T_p| - N_{\text{divisible}} = p^3 - p^2$

This corresponds to Option D.