Number of Symmetric or Skew Symmetric Matrices with Determinant Divisible by Prime

To find the number of matrices $A \in T_p$ such that $A$ is symmetric, skew-symmetric, or both, and $\det(A)$ is divisible by $p$, we analyze the conditions step-by-step.

Given:

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

The determinant of $A$ is:
$\det(A) = a^2 - bc$

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Step 1: Condition for Skew-Symmetry

A matrix $A$ is skew-symmetric if $A^T = -A$.

$$\begin{bmatrix} a & c \\ b & a \end{bmatrix} = \begin{bmatrix} -a & -b \\ -c & -a \end{bmatrix}$$

This requires:

1. $a = -a \implies a = 0$
2. $b = -c \implies b + c = 0$

Since $b, c \in \{0, 1, 2, \dots, p-1\}$, the non-negativity of entries implies $b = 0$ and $c = 0$.

Thus, the only skew-symmetric matrix in $T_p$ is the zero matrix:

$$A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

For this matrix, $\det(A) = 0$, which is divisible by $p$. Note that this matrix is also symmetric (since $b = c = 0$).

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Step 2: Condition for Symmetry

A matrix $A$ is symmetric if $A^T = A$, which implies $b = c$.

Substitute $c = b$ into the matrix and its determinant:

$$A = \begin{bmatrix} a & b \\ b & a \end{bmatrix}, \quad \det(A) = a^2 - b^2 = (a - b)(a + b)$$

We are given that $\det(A)$ is divisible by $p$:
$p \mid (a - b)(a + b)$

Since $p$ is an odd prime, by Euclid's lemma, $p$ must divide at least one of the factors:
$p \mid (a - b) \quad \text{or} \quad p \mid (a + b)$

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Step 3: Counting the Pairs $(a, b)$

1. Case 1: $p \mid (a - b)$
   Since $a, b \in \{0, 1, 2, \dots, p-1\}$, we have:
   $-(p - 1) \le a - b \le p - 1$
   The only multiple of $p$ in this range is $0$. Thus:
   $a - b = 0 \implies a = b$
   Since $a$ can be any integer from $0$ to $p - 1$, there are $p$ such pairs:
   $(0, 0), (1, 1), (2, 2), \dots, (p-1, p-1)$

2. Case 2: $p \mid (a + b)$
   Since $a, b \in \{0, 1, 2, \dots, p-1\}$, we have:
   $0 \le a + b \le 2p - 2$
   The possible multiples of $p$ in this range are $0$ and $p$.

   • Subcase 2a: $a + b = 0 \implies a = 0, b = 0$. This gives the pair $(0, 0)$, which is already counted in Case 1.
   • Subcase 2b: $a + b = p$.
     The possible pairs $(a, b)$ are:
     $(1, p-1), (2, p-2), \dots, (p-1, 1)$
     There are $p - 1$ such pairs.

3. Check for Overlap between Case 1 and Subcase 2b:
   If a pair satisfies both $a = b$ and $a + b = p$, then:
   $2a = p$
   Since $p$ is an odd prime, $2a = p$ has no integer solutions. Hence, the set of pairs where $a = b$ and the set where $a + b = p$ are completely disjoint.

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Step 4: Total Count

The total number of symmetric matrices (which includes the single skew-symmetric matrix) is:
$\text{Total number of matrices} = p + (p - 1) = 2p - 1$

Thus, the correct option is D.