Question

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}{*{20}{c}} a&b; \\\ c&a; \end{array}} \right]:a,b,c \in \left\\{ {0,1,2.......\left( {p - 1} \right)} \right\\}} \right\\}
The number of A in ${T_p}$ such that the trace of A is not divisible by p but det (A) is divisible by p is?
[Note: The trace of a matrix is the sum of its diagonal entries]
$\left( a \right)\left( {p - 1} \right)\left( {{p^2} - p + 1} \right)$
$\left( b \right){p^2} - {\left( {p - 1} \right)^2}$
$\left( c \right){\left( {p - 1} \right)^2}$
$\left( d \right)\left( {p - 1} \right)\left( {{p^2} - 2} \right)$

Answer 1

In this particular type of question use the concept that if 0 is divided by something then it is zero so we cannot take value 0 if we want some remainder and use the concept that if there are n object so the number of ways to select any one object is given as ${}^n{C_1} = n$ so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
{T_p} = \left\\{ {A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&a; \end{array}} \right]:a,b,c \in \left\\{ {0,1,2.......\left( {p - 1} \right)} \right\\}} \right\\}, Where p is an odd prime number.
Now we have to find out the number of A in ${T_p}$ such that the trace of A is not divisible by p but det (A) is divisible by p.
Since trace of A i.e. trace of matrix A is the sum of diagonal elements.
So the trace of A is, ${\text{tr}}\left( A \right) = 2a$
Now according to the given condition trace cannot be divisible by P therefore, $a \ne 0$ otherwise it is divisible by P.
Now the number of ways to select a as a \in \left\\{ {0,1,2.......\left( {p - 1} \right)} \right\\} but a cannot be zero so the number of ways to select a are (p – 1) ways.
Let the remainder when we divide this a by p is r.
i.e. $\dfrac{a}{p} = q\dfrac{r}{p}$, where q is quotient, r is remainder and p is divisor.
Now the determinant of A is $\left| A \right| = {a^2} - bc$
Now it is given that this determinant is divisible by p.
So, if $\dfrac{{{a^2}}}{p}$ and $\dfrac{{bc}}{p}$ has same remainder when divisible by p then ${a^2} - bc$ is divisible by p.
Now the number of ways to select a, we calculated earlier.
Now b is multiplied by c and both are belongs to b,c \in \left\\{ {0,1,2.......\left( {p - 1} \right)} \right\\} now we cannot take b and c equal to zero otherwise it is completely divisible by p.
So if we choose b, the number of ways is (p – 1) so there is one unique value of c such that the product of bc when divided by p gives r as remainder.
Now as ${a^2}{\text{ and }}bc$ has the same remainder r when divided by p, so ${a^2} - bc$ is divisible by p.
Therefore a and b can be taken in (p – 1) ways and c can be chosen in only one way.
So the total number of A in ${T_p}$ such that the trace of A is not divisible by p but det (A) is divisible by p is,
$\Rightarrow {\left( {p - 1} \right)^2} \times 1$
So the total number of A is ${\left( {p - 1} \right)^2}$
So this is the required answer.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that if an equation is an addition or subtraction of two or more than two terms and if both the terms when divided by some number and leaves the same remainder then the complete equation i.e. addition or subtraction of two or more terms are completely divisible by the same number.