JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Matrices

For $A=\left[\begin{array}{cc}a& b\\ c& a\end{array}\right]$ to be symmetric, we need $b=c$. For skew-symmetric, $a=0$ and $b=-c$, which in modulo $p$ means $b+c\equiv 0(\mathrm{mod}p)$. The determinant is $\det (A)=a^2-bc$. For $\det (A)$ divisible by $p$: $a^2\equiv bc(\mathrm{mod}p)$.

Symmetric case ($b=c$): $a^2\equiv b^2(\mathrm{mod}p)$, so $(a-b)(a+b)\equiv 0(\mathrm{mod}p)$. This gives $a\equiv b$ or $a\equiv -b(\mathrm{mod}p)$. Each yields $p$ choices (including $a=b=0$), but we subtract 1 for overlap at $a=b=0$, giving $2p-1$ symmetric matrices.

Skew-symmetric case ($a=0$, $b+c\equiv 0$): $\det (A)=-bc\equiv 0$ automatically when $b+c\equiv 0(\mathrm{mod}p)$. This gives $p$ choices for $b$ (then $c=-b$), all satisfying the determinant condition.

However, the zero matrix ($a=b=c=0$) is both symmetric and skew-symmetric. Counting carefully: symmetric with divisible determinant gives $2p-1$ matrices (this already includes the zero matrix). The answer is $2p-1$.