Question

Let $p$ be an odd prime number and ${{\text{T}}_p}$ be the following set of ${\text{2}} \times {\text{2}}$ matrices:
${T_p} = \left\{ {A{\text{ = }}\left[ {\left( {a{\text{,}}b} \right),\left( {c{\text{,}}a} \right)} \right]:a{\text{,}}b{\text{,}}c \in \left\{ {0,1,....,p - {\text{1}}} \right\}} \right\}$
The number of $A$ in ${T_p}$ such that $A$ is either symmetric or skew-symmetric or both,
and det$\left( A \right)$ divisible by $p$ is:
$
  {\text{A}}{\text{. }}{\left( {p - 1} \right)^2} \\
  {\text{B}}{\text{. 2}}\left( {p - 1} \right) \\
  {\text{C}}{\text{. }}{\left( {p - 1} \right)^2} + 1 \\
  {\text{D}}{\text{. 2}}p - 1 \\
 $

Answer 1

Hint: First, we will find the determinants of this matrix as in the case of symmetric and skew-symmetric. Then, we will find the values of all these determinants which can be divisible by $p$ and add both the answers of symmetric as well as skew-symmetric matrix.

Complete step-by-step answer:
Given $A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&a
\end{array}} \right]$
As you know, Determinant can be computed as for example $\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc$
Therefore, by apply this method we will get,
$\left| A \right| = {a^2} - bc$
If $A$ is symmetric then $b = c$
So $\left| A \right| = {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ which is divisible by $p$, if $\left( {a + b} \right)$ is divisible by $p$ or $\left( {a - b} \right)$ is divisible by $p.$
Now $\left( {a + b} \right)$ is divisible by $p$ if $\left( {a,b} \right)$ can take values $\left( {1,p - 1} \right),\left( {2,p - 2} \right),\left( {3,p - 3} \right)......\left( {p - 1,1} \right)$.
Therefore,$\left( {p - 1} \right)$ ways.
Also $\left( {a - b} \right)$ is divisible by $p$ only when $a - b = 0$ , that is,$a = b$. Then $\left( {a,b} \right)$ can take values$\left( {0,0} \right),\left( {1,1} \right),\left( {2,2} \right)....\left( {p - 1,p - 1} \right)$ .Therefore,$p$ ways
If $A$ is skew-symmetric then $a = 0$ and $b = - c$ or $b + c = 0$ which gives
 $\left| A \right| = 0$ when ${b^2} = 0$ $ \Rightarrow b = 0,c = 0$.
But this possibility is already included when
$A$ is symmetric and $\left( {a,b} \right) = \left( {0,0} \right)$
Again if $A$ is both symmetric and skew-symmetric, then clearly $A$ is a null matrix which case is already included. Hence total number of ways $ = p + p - 1 = 2p - 1$.

Note: Here, in this question one should consider all the cases for the value of the determinant of $A$ as this can be a common mistake done by students. For this question we have to first understand the concepts of symmetric and skew-symmetric matrices and determinants i.e.
 The non-diagonal elements of a symmetric matrix are equal whereas the elements on the diagonal of a skew-symmetric matrix are zero.