Let p be an odd prime number and $T_p$ be the following set of matrices $T_p=\left\{A=\begin{bmatrix}a&b\\c&a\end{bmatrix}: a, b, c ∈ \{0, 1, 2,..., p-1\}\right\}$

The number of A in $T_p$ such that the trace of A is not divisible by p but det (A) is divisible by p, is

$(p-1)^2$

We have, trace of $A = 2a$.

Clearly, trace of A will be divisible by p iff $a = 0$.

$∴a∈\{1, 2,..., (p-1)\}$

We have, $det (A) = a^2-bc$

$∴p❘ det\, (A)$

$⇒p|a^2-bc$

$⇒a^2-bc = λp$ for some integer λ

For each value of $a ∈ \{1, 2,..., (p-1)\}$, there are $(p-1)$ ordered pairs $(b, c)$ for bc such that $a^2 -bc$ is divisible by p.

Hence, there are $(p-1)^2$ matrices in $T_p$ such that trace of A is not divisible by p but determinant of A is divisible by p.