Question

Let $p$ be an odd prime number and $T_p$ be the following set of $2\times 2$ matrices:

$T_p=\{A=\left[\begin{array}{cc}a& b\\ c& a\end{array}\right]:a,b,c\in \{0,1,\dots .,p-1\left\}\right\}$

The number of A in $T_p$ such that det(A) is not divisible by p is

• A. $2p^2$
• B. $p^3-5p$
• C. $p^3-3p$
• D. $p^3-p^2$

Answer 1

Answer: (D)

$p^3-p^2$

Given $A=\left[\begin{array}{cc}a& b\\ c& a\end{array}\right]:a,b,c\in \{0,1,\dots .,p-1\}\}$

$|A|=a^2-bc$

We will find number of $A$ in $T_p$ such that A is either symmetric or skew-symmetric or neither and $|A|$ is divisible by $p$

Case I : If $A$ is symmetric
Then $b=c$
$\Rightarrow |A|=a^2-b^2$

Since, $|A|$ is divisible by $p$.
$\Rightarrow a^2-b^2$ is divisible by $p$.
$\Rightarrow (a-b)(a+b)$ is divisible by $p$
Now, $a-b$ is divisible by $p$ if $a=b$ , then $p$ cases are possible
and $a+b$ is divisible by $p$ if $a+b=0$ or $a+b=p$
For $a+b=0$, the only solution is $a=0,b=0$ which is already counted.
Hence, for $a+b=p$ , we have p-1 cases possible as a can take any value from $1,2...p-1$
So, total cases $=p+p-1=2p-1$
Case II: If A is skew-symmetric
Then $a=0,b+c=0$
$\Rightarrow |A|=b^2$
Since, |A| is divisible by p
$b^2$ is divisible by p which is possible only when $b=0$
When $b=0$, A will be a zero matrix which is already counted in symmetric matrix.
Hence, no cases possible.
So, 2p-1 matrices are divisible by $p$
Now, we will find number of A in $T_p$ divisible by p when $a\ne 0$,
Since, $a\ne 0$ , so values of a can be chosen in p-1 ways.
For each such fixed $a$ we need to determine all ordered pairs (b, c) for which $a^2-bc$ is divisible by $p$.
Let $r$ be the remainder when $a^2$ is divided by $p$
For $a^2-bc$ to be divisible by $p$ is equivalent to saying that the integer $bc$ also leaves the remainder $r$ when divided by $p$.
Hence, $b$ cannot be 0
So, $b$ can be chosen in $p-1$ ways and $c$ in only one way.
Hence, number of matrices A in $T_p$ are $(p-1{)}^2$ when $a\ne 0$
Now, since every matrix in $T_p$ is determined by three mutually independent parameters $a,b,c$ each taking $p$ possible values. So, in all there are $p^3$ matrices in the set $T_p$.
Number of A in $T_p$ which is not divisible by $p$ = Total cases - ($|A|$ divisible by p, when $a\ne 0$) -($|A|$ divisible by p, when $a=0$)
$=p^3-(p-1{)}^2-(2p-1)$
$=p^3-p^2$