Question

The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and det($A$) divisible by $p$ is:

• A. $(p-1{)}^2$
• B. $2(p-1)$
• C. $(p-1{)}^2+1$
• D. $2p-1$

Answer 1

The correct option is D $2p-1$

For a matrix $A$ to be symmetric, $A=A^T$, which is possible if $b=c$.
Hence we get $|A|=a^2-b^2$
We must have $a^2-b^2=kp$
$\Rightarrow (a+b)(a-b)=kp$
$\Rightarrow$ either $a-b$ or $a+b$ is a multiple of $p$
$a-b$ will be divisible by $p$ only when $a=b$ (since $a$ and $b$ can take values from $0$ to $p-1$); hence number of such matrices is $p$
and when $a+b=$ multiple of p $\Rightarrow a,b$ can take $p-1$ values. (the pairs $(1,p-1),(2,p-2),(3,p-3)$ and so on)
$\therefore$Total number of matrices $=p+p-1$
$=2p-1.$
If a matrix $A$ is skew symmetric then $A=-A^T$, which, in this case, is possible only if $a=0$ and $b=0$. That gives a null matrix which has already been counted once in skew symmetric case.
Hence, total number of matrix is still $2p-1$