Let p be an odd prime number and T_{p} be the | vedclass

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(B) $1.$ For $A$ to be symmetric,$b=c$. For $A$ to be skew-symmetric,$a=0$ and $b=-c$. If $A$ is symmetric,$A = \begin{bmatrix} a & b \ b & a \end{bm

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$(D, C, D)$

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(B) $1.$ For $A$ to be symmetric,$b=c$. For $A$ to be skew-symmetric,$a=0$ and $b=-c$. If $A$ is symmetric,$A = \begin{bmatrix} a & b \ b & a \end{bmatrix}$,$\det(A) = a^2-b^2$. $\det(A) \equiv 0 \pmod{p} \implies a^2 \equiv b^2 \pmod{p} \implies a \equiv \pm b \pmod{p}$. For $a=b$,there are $p$ choices. For $a=-b$,there are $p$ choices. Excluding the case $a=b=0$ (counted twice),we have $2p-1$ symmetric matrices. If $A$ is skew-symmetric,$a=0, b=-c$. $\det(A) = 0 - b(-b) = b^2$. $\det(A) \equiv 0 \pmod{p} \implies b=0$. Thus $A$ is the zero matrix,which is already counted. Total count is $2p-1$. Correct option is $(D)$.$2.$ $	ext{tr}(A) = 2a$. Since $p$ is an odd prime,$2a 
ot\equiv 0 \pmod{p} \implies a 
ot\equiv 0 \pmod{p}$. There are $p-1$ choices for $a$. $\det(A) = a^2 - bc \equiv 0 \pmod{p} \implies bc \equiv a^2 \pmod{p}$. For each $a \in \{1, \ldots, p-1\}$,$a^2 
ot\equiv 0 \pmod{p}$. Thus $b$ can be any of $p-1$ values (excluding $0$),and $c$ is uniquely determined as $c \equiv a^2 b^{-1} \pmod{p}$. Total count is $(p-1)(p-1) = (p-1)^2$. Correct option is $(C)$.$3.$ $\det(A) = a^2 - bc 
ot\equiv 0 \pmod{p}$. Total matrices in $T_p$ is $p^3$. If $a=0$,$\det(A) = -bc 
ot\equiv 0 \implies b 
eq 0, c 
eq 0$. Choices: $(p-1)(p-1) = (p-1)^2$. If $a 
eq 0$,$bc 
ot\equiv a^2 \pmod{p}$. For each $a$,there are $p-1$ pairs $(b, c)$ such that $bc \equiv a^2 \pmod{p}$. Total pairs $(b, c)$ is $p^2$. So $p^2 - (p-1)$ pairs satisfy $\det(A) 
ot\equiv 0$. Total = $(p-1)^2 + (p-1)(p^2 - p + 1) = (p-1)(p-1 + p^2 - p + 1) = (p-1)p^2 = p^3 - p^2$. Correct option is $(D)$.

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