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Let $p$ be an odd prime number and $T_p$ be the following set of $2 \times 2$ matrices
$$
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
$$Question:
The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and det $(A)$ is divisible by $p$ is
Options:
Let $p$ be an odd prime number and $T_p$ be the following set of $2 \times 2$ matrices
$$
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
$$Question:
The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and det $(A)$ is divisible by $p$ is
Solution:
1273 Upvotes
Verified Answer
The correct answer is:
$2 p-1$
$2 p-1$
Given, $A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]$,
$a, b, c \in\{0,1,2, \ldots, p-1\}$
If $A$ is skew-symmetric matrix, then $a=0, b=-c$
$\therefore \quad|A|=-b^2$.
Thus, $P$ divides $|A|$ only when $b=0$...(i)
Again, if $A$ is symmetric matrix, then $b=c$ and $|A|=a^2-b^2$.
Thus, $p$ divides $|A|$ if either $p$ divides $(a-b)$ or $p$ divides $(a+b)$. $p$ divides $(a-b)$, only when $a=b$ ie, $a=b \in\{0,1,2, \ldots,(p-1)\}$
ie, pchoices
$p$ divides $(a+b)$.
$\Rightarrow p$ choices, including $a=b=0$ included in (i)
$\therefore$ Total number of choices are $(p+p-1)=2 p-1$.
Hence, (c) is the correct option.
$a, b, c \in\{0,1,2, \ldots, p-1\}$
If $A$ is skew-symmetric matrix, then $a=0, b=-c$
$\therefore \quad|A|=-b^2$.
Thus, $P$ divides $|A|$ only when $b=0$...(i)
Again, if $A$ is symmetric matrix, then $b=c$ and $|A|=a^2-b^2$.
Thus, $p$ divides $|A|$ if either $p$ divides $(a-b)$ or $p$ divides $(a+b)$. $p$ divides $(a-b)$, only when $a=b$ ie, $a=b \in\{0,1,2, \ldots,(p-1)\}$
ie, pchoices
$p$ divides $(a+b)$.
$\Rightarrow p$ choices, including $a=b=0$ included in (i)
$\therefore$ Total number of choices are $(p+p-1)=2 p-1$.
Hence, (c) is the correct option.
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