JEE Advanced Maths PYQ – Determinants: Minor & Co-factor, Value of a Determinant

Q 1 :

Let $ω \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $[\begin{matrix} 1 & a & b \\ ω & 1 & c \\ ω^{2} & ω & 1 \end{matrix}]$ where each of $a$ , $b$ and $c$ is either $ω$ or $ω^{2}$ . Then the number of distinct matrices in the set $S$ is [2011]

2
6
4
8

1

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3

4

(1)

For the given matrix to be non-singular

$| \begin{matrix} 1 & a & b \\ ω & 1 & c \\ ω^{2} & ω & 1 \end{matrix} |$ $\neq 0$

$\Rightarrow 1 - (a + c) ω + a c ω^{2} \neq 0$ $\Rightarrow (1 - a ω) (1 - c ω) \neq 0$

$\Rightarrow a \neq ω^{2} and c \neq ω^{2},$ where $ω$ is a complex cube root of unity.

As $a$ , $b$ and $c$ are complex cube roots of unity,

$∴$ $a$ and $c$ can take only one value, i.e. $ω$ , while $b$ can take two values, i.e. $ω$ and $ω^{2}$ .

$∴$ Total number of distinct matrices in the set S $= 1 \times 1 \times 2 = 2$

Q 2 :

Consider three points

$P = (- \sin (β - α), - \cos β), Q = (\cos (β - α), \sin β)$ and

$R = (\cos (β - α + θ), \sin (β - θ)),$ where $0 < α, β, θ < \frac{π}{4} .$

Then, [2008]

P lies on the line segment RQ
Q lies on the line segment PR
R lies on the line segment QP
P, Q, R are non-collinear

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(4)

$Given: Three points P (- \sin (β - α), - \cos β), Q (\cos (β - α), \sin β)$

$and R (\cos (β - α + θ), \sin (β - θ)),$

where $0 < α, β, θ < \frac{π}{4}$

$∴$ $Δ =$ $| \begin{matrix} 1 & 1 & 1 \\ - \sin (β - α) & \cos (β - α) & \cos (β - α + θ) \\ - \cos β & \sin β & \sin (β - θ) \end{matrix} |$ $[C_{3} \to C_{3} - (C_{1} \sin θ + C_{2} \cos θ)]$

$Δ =$ $| \begin{matrix} 1 & 1 & 1 - \sin θ - \cos θ \\ - \sin (β - α) & \cos (β - α) & 0 \\ - \cos β & \sin β & 0 \end{matrix} |$

$= (1 - \sin θ - \cos θ) [\cos β \cos (β - α) - \sin β \sin (β - α)]$

$= [1 - (\sin θ + \cos θ)] \cos (2 β - α)$

$∵$ $0 < α, β, θ < \frac{π}{4},$ $∴$ $\sin θ + \cos θ \neq 1$

Also, $2 β - α < \frac{π}{2} \Rightarrow \cos (2 β - α) \neq 0$

$∴$ $Δ \neq 0 \Rightarrow$ The three given points are non-collinear.

Q 3 :

Let $S = {A = (\begin{matrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{matrix}) : a, b, c, d, e \in {0, 1} and | A | \in {- 1, 1}},$ where $| A |$ denotes the determinant of $A$ .

Then the number of elements in $S$ is ______________. [2024]

(16)

$| A |$ $= - (e - d) + c (b - a) = \pm 1$

Case (i): $c = 0 \Rightarrow e - d = \pm 1$

$\Rightarrow (e, d) = (1, 0), (0, 1) \to 2 ways$

$b$ and $a$ can each take $2$ ways

$\Rightarrow Total = 1 \times 2 \times 2 \times 2 = 8 ways$

Case (ii): $c = 1 \Rightarrow d - e + b - a = \pm 1$

$\begin{matrix} c & a & b & d & e \\ 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{matrix}}$ $\Rightarrow 4 \times 2 = 8 ways$

$∴$ $Total = 16 ways$

Q 4 :

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is 3 and the trace of $A^{3}$ is $- 18$ , then the value of the determinant of $A$ is _____. [2020]

(5)

$∵$ $The trace of A is 3$

Let $A = [\begin{matrix} x & y \\ z & 3 - x \end{matrix}];$ Now, $A^{2} = [\begin{matrix} x & y \\ z & 3 - x \end{matrix}] [\begin{matrix} x & y \\ z & 3 - x \end{matrix}]$

$= [\begin{matrix} x^{2} + y z & x y + 3 y - x y \\ x z + 3 z - x z & y z + 9 + x^{2} - 6 x \end{matrix}]$ $= [\begin{matrix} x^{2} + y z & 3 y \\ 3 z & y z + 9 + x^{2} - 6 x \end{matrix}]$

$\Rightarrow A^{3} = [\begin{matrix} x^{2} + y z & 3 y \\ 3 z & y z + 9 + x^{2} - 6 x \end{matrix}] [\begin{matrix} x & y \\ z & 3 - x \end{matrix}]$

$= [\begin{matrix} x^{3} + x y z + 3 y z & x^{2} y + y^{2} z + 9 y - 3 x y \\ 3 z x + y z^{2} + 9 z + x^{2} z - 6 x z & 6 y z + 27 + 3 x^{2} - 18 x - x y z - 9 x - x^{3} + 6 x^{2} \end{matrix}]$

Given that the trace of $A^{3}$ is $- 18$

$∴$ $x^{3} + x y z + 3 y z + 6 y z + 27 + 3 x^{2} - 18 x - x y z - 9 x - x^{3} + 6 x^{2} = - 18$

$\Rightarrow 9 y z + 9 x^{2} - 27 x + 27 = - 18$ $\Rightarrow y z + x^{2} - 3 x + 3 = - 2$

$\Rightarrow 3 x - x^{2} - y z = 5$ $\dots (i)$

Now, $| A |$ $= 3 x - x^{2} - y z$

Hence, $| A |$ $= 5$ $from (i)$

Q 5 :

Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

(4)

$Let Det (P) =$ $| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} |$

$= a_{1} (b_{2} c_{3} - b_{3} c_{2}) - a_{2} (b_{1} c_{3} - b_{3} c_{1}) + a_{3} (b_{1} c_{2} - b_{2} c_{1})$

If $a_{1} = 1,$ $a_{2} = - 1,$ $a_{3} = 1,$ $b_{2} c_{3} = b_{1} c_{3} = b_{1} c_{2} = 1$ and $b_{3} c_{2} = b_{3} c_{1} = b_{2} c_{1} = - 1$

then the maximum value of $Det (P) = 6$

But it is not possible, as

$(b_{2} c_{3}) (b_{3} c_{1}) (b_{1} c_{2}) = - 1$ and $(b_{1} c_{3}) (b_{3} c_{2}) (b_{2} c_{1}) = 1,$

i.e., $b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = 1$ and $b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = - 1$

This is a contradiction.

Similarly, contradiction occurs when

$a_{1} = 1,$ $a_{2} = 1,$ $a_{3} = 1,$ $b_{2} c_{3} = b_{3} c_{1} = b_{1} c_{2} = 1$ and $b_{3} c_{2} = b_{1} c_{3} = b_{2} c_{1} = - 1$

Now, for the value to be 5, one of the terms must be zero, but that will make two terms zero, which means the answer cannot be 5.

Now $| \begin{matrix} 1 & 1 & 1 \\ - 1 & 1 & 1 \\ 1 & - 1 & 1 \end{matrix} |$ $= 4;$

Therefore, the maximum value is 4

Q 6 :

Let $z = \frac{- 1 + \sqrt{3} i}{2},$ where $i = \sqrt{- 1}$ , and $r, s \in {1, 2, 3}$ . Let $P = [\begin{matrix} {(- z)}^{r} & z^{2 s} \\ z^{2 s} & z^{r} \end{matrix}]$ and $I$ be the identity matrix of order 2. Then the total number of ordered pairs $(r, s)$ for which $P^{2} = - I$ is [2016]

(1)

$z = \frac{- 1 + i \sqrt{3}}{2} \Rightarrow z^{3} = 1 and 1 + z + z^{2} = 0$

$P^{2} = [\begin{matrix} {(- z)}^{r} & z^{2 s} \\ z^{2 s} & z^{r} \end{matrix}] [\begin{matrix} {(- z)}^{r} & z^{2 s} \\ z^{2 s} & z^{r} \end{matrix}]$

$= [\begin{matrix} z^{2 r} + z^{4 s} & z^{2 s} ({(- z)}^{r} + z^{r}) \\ z^{2 s} ({(- z)}^{r} + z^{r}) & z^{4 s} + z^{2 r} \end{matrix}]$

For $P^{2} = - I,$ we should have

$z^{2 r} + z^{4 s} = - 1$ and $z^{2 s} ({(- z)}^{r} + z^{r}) = 0$

$\Rightarrow z^{2 r} + z^{4 s} + 1 = 0$ and ${(- z)}^{r} + z^{r} = 0$

$\Rightarrow r is odd and s = r but not a multiple of 3,$

which is possible when $s = r = 1$

$∴$ only one pair is there.

Q 7 :

Let $I = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ and $P = (\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}) .$ Let $Q = (\begin{matrix} x & y \\ z & 4 \end{matrix})$ for some non-zero real numbers $x$ , $y$ and $z$ , for which there is a $2 \times 2$ matrix $R$ with all entries being non-zero real numbers, such that $Q R = R P .$

Then which of the following statement(s) is (are) TRUE? [2025]

The determinant of $Q - 2 I$ is zero
The determinant of $Q - 6 I$ is 12
The determinant of $Q - 3 I$ is 15
$y z = 2$

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4

Select one or more options

(1, 2)

$Given: Q R = R P \Rightarrow Q R - 2 R = R P - 2 R$

$\Rightarrow$ $| Q - 2 I |$ $| R |$ $=$ $| R |$ $| \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} |$ $=$ $| R |$ $\cdot 0$

$\Rightarrow$ $| Q - 2 I |$ $= 0$ $\dots (i)$

Also, $Q R - 6 R = R P - 6 R$

$\Rightarrow$ $| Q - 6 I |$ $| R |$ $=$ $| R |$ $| \begin{matrix} - 4 & 0 \\ 0 & - 3 \end{matrix} |$ $=$ $| R |$ $\cdot 12$

So, $| Q - 6 I |$ $= 12$ $\dots (i i)$

Now, $Q R - 3 R = R P - 3 R$

$\Rightarrow$ $| Q - 3 I |$ $| R |$ $=$ $| R |$ $| \begin{matrix} - 1 & 0 \\ 0 & 0 \end{matrix} |$ $=$ $| R |$ $\cdot 0$

So, $| Q - 3 I |$ $= 0$ $\dots (i i i)$

Now, from (i), $| Q - 2 I |$ $= 0$

$\Rightarrow$ $| \begin{matrix} x - 2 & y \\ z & 2 \end{matrix} |$ $= 0$ $\Rightarrow 2 x - 4 = y z$ $\dots (i v)$

From (iii), $| Q - 3 I |$ $= 0$

$\Rightarrow$ $| \begin{matrix} x - 3 & y \\ z & 1 \end{matrix} |$ $= 0$ $\Rightarrow x - 3 = y z$ $\dots (v)$

From (iv) and (v), $2 x - 4 = x - 3$

$\Rightarrow x = 1,$

so, $y z = - 2$

Q 8 :

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N = N M .$ Further, if $M \neq N^{2}$ and $M^{2} = N^{4},$ then [2014]

determinant of $(M^{2} + M N^{2})$ is 0
There is a $3 \times 3$ non-zero matrix $U$ such that $(M^{2} + M N^{2}) U$ is the zero matrix
determinant of $(M^{2} + M N^{2}) \geq 1$
For a $3 \times 3$ matrix $U$ , if $(M^{2} + M N^{2}) U$ equals the zero matrix, then $U$ is the zero matrix

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4

Select one or more options

(1, 2)

$Given: M N = N M, M \neq N^{2} and M^{2} = N^{4}$

Then, $M^{2} = N^{4}$ $\Rightarrow (M + N^{2}) (M - N^{2}) = 0$

$\Rightarrow (i) M + N^{2} = 0 and M - N^{2} \neq 0,$

$(ii)$ $| M + N^{2} |$ $\neq 0 and$ $| M - N^{2} |$ $= 0$

In each case, $| M + N^{2} |$ $= 0$

$∴$ $| M^{2} + M N^{2} |$ $=$ $| M |$ $| M + N^{2} |$ $= 0$

$∴$ (1) is correct and (3) is not correct.

Also, we know if $| A |$ $= 0,$ then there can be many matrices $U$ , such that $A U = 0$

$∴$ $(M^{2} + M N^{2}) U = 0$ will be true for many values of $U$ .

$∴$ (2) is correct.

Again, if $A X = 0$ and $| A |$ $= 0,$ then $X$ can be non-zero.

$∴$ (4) is not correct.

Q 9 :

Consider the lines given by

$L_{1} : x + 3 y - 5 = 0,$ $L_{2} : 3 x - k y - 1 = 0,$ $L_{3} : 5 x + 2 y - 12 = 0$

Match the Statements / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [2008]

Column I Column II

(A) $L_{1}, L_{2}, L_{3}$ are concurrent, if (p) $k = - 9$

(B) One of $L_{1}, L_{2}, L_{3}$ is parallel to at least one of the other two, if (q) $k = - \frac{6}{5}$

(C) $L_{1}, L_{2}, L_{3}$ form a triangle, if (r) $k = \frac{5}{6}$

(D) $L_{1}, L_{2}, L_{3}$ do not form a triangle, if (s) $k = 5$

(A) → (s); (B) → (p, q); (C) → (r); (D) → (p, q, s)
(A) → (p, q, s); (B) → (p, q); (C) → (r); (D) → (s)
(A) → (p, q, s); (B) → (r); (C) → (p, q); (D) → (s)
(A) → (r); (B) → (p, q, s); (C) → (p, q); (D) → (s)

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(1)

(A) → (s); (B) → (p, q); (C) → (r); (D) → (p, q, s)

The given lines are

$L_{1} : x + 3 y - 5 = 0$

$L_{2} : 3 x - k y - 1 = 0$

$L_{3} : 5 x + 2 y - 12 = 0$

(A) Three lines $L_{1}, L_{2}, L_{3}$ are concurrent, if

$| \begin{matrix} 1 & 3 & 5 \\ 3 & - k & 1 \\ 5 & 2 & 12 \end{matrix} |$ $= 0$ $\Rightarrow 13 k - 65 = 0$ $\Rightarrow k = 5$

$∴$ $(A) \to (s)$

(B) For $L_{1} ∥ L_{2}$ $\Rightarrow \frac{1}{3} = \frac{- 3}{k}$ $\Rightarrow k = - 9$

and $L_{2} ∥ L_{3}$ $\Rightarrow \frac{3}{5} = \frac{- k}{2}$ $\Rightarrow k = - \frac{6}{5}$

$∴$ $(B) \to (p), (q)$

(C) Three lines $L_{1}, L_{2}, L_{3}$ will form a triangle if no two of them are parallel and no three are concurrent.

$∴$ $k \neq 5, - 9, - \frac{6}{5}$

$∴$ $(C) \to (r)$

(D) Three lines $L_{1}, L_{2}, L_{3}$ do not form a triangle if either any two of these are parallel or the three are concurrent, i.e.,

$k = 5, - 9, - \frac{6}{5}$

$∴$ $(D) \to (p), (q), (s)$

Q 10 :

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

$T_{p} = {A = [\begin{matrix} a & b \\ c & a \end{matrix}] : a, b, c \in {0, 1, 2, \dots, p - 1}}$ [2010]

Q. 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

${(p - 1)}^{2}$
$2 (p - 1)$
${(p - 1)}^{2} + 1$
$2 p - 1$

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4

(4)

$Given, A = [\begin{matrix} a & b \\ c & a \end{matrix}], a, b, c \in {0, 1, 2, \dots, p - 1}$

If $A$ is a skew-symmetric matrix, then $a = 0,$ $b = - c$

$∴$ $| A |$ $= - b^{2}$

Thus, $p$ divides $| A |$ , only when $b = 0 .$ $\dots (i)$

Again, if $A$ is a 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 $a = b,$

i.e., $a = b \in {0, 1, 2, \dots, (p - 1)},$

i.e., $p$ choices. $\dots (i i)$

$p$ divides $(a + b)$ .

$\Rightarrow p$ choices, including $a = b = 0$ included in Eq. (i).

$∴$ Total number of choices are $(p + p - 1) = 2 p - 1$

Q 11 :

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

$T_{p} = {A = [\begin{matrix} a & b \\ c & a \end{matrix}] : a, b, c \in {0, 1, 2, \dots, p - 1}}$ [2010]

Q. 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.]

$(p - 1) (p^{2} - p + 1)$
$p^{3} - {(p - 1)}^{2}$
${(p - 1)}^{2}$
$(p - 1) (p^{2} - 2)$

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3

4

(3)

Trace of $A = 2 a$ , will not be divisible by $p$ , if $a \neq 0 .$

$| A |$ $= a^{2} - b c,$ for $(a^{2} - b c)$ to be divisible by $p$ , there are exactly $(p - 1)$ ordered pairs $(b, c)$ for any value of $a$ .

$∴$ Required number is ${(p - 1)}^{2}$

Q 12 :

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

$T_{p} = {A = [\begin{matrix} a & b \\ c & a \end{matrix}] : a, b, c \in {0, 1, 2, \dots, p - 1}}$ [2010]

Q. The number of $A$ in $T_{p}$ such that $\det (A)$ is not divisible by $p$ is

$2 p^{2}$
$p^{3} - 5 p$
$p^{3} - 3 p$
$p^{3} - p^{2}$

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2

3

4

(4)

The number of matrices for which $p$ does not divide $Tr (A)$ $= (p - 1) p^{2}$ of these, ${(p - 1)}^{2}$ are such that $p$ divides $| A |$ . The number of matrices for which $p$ divides $Tr (A)$ and $p$ does not divide $| A |$ are ${(p - 1)}^{2} .$

$∴$ Required number $= (p - 1) p^{2} - {(p - 1)}^{2} + {(p - 1)}^{2} = p^{3} - p^{2}$

Topic Question Set

Q 2 :

Consider three points

$P = (- \sin (β - α), - \cos β), Q = (\cos (β - α), \sin β)$ and

$R = (\cos (β - α + θ), \sin (β - θ)),$ where $0 < α, β, θ < \frac{π}{4} .$

Then, [2008]

Q 3 :

Let $S = {A = (\begin{matrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{matrix}) : a, b, c, d, e \in {0, 1} and | A | \in {- 1, 1}},$ where $| A |$ denotes the determinant of $A$ .

Then the number of elements in $S$ is ______________. [2024]

Q 4 :

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is 3 and the trace of $A^{3}$ is $- 18$ , then the value of the determinant of $A$ is _____. [2020]

Q 5 :

Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

Q 8 :

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N = N M .$ Further, if $M \neq N^{2}$ and $M^{2} = N^{4},$ then [2014]

Q 12 :

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

$T_{p} = {A = [\begin{matrix} a & b \\ c & a \end{matrix}] : a, b, c \in {0, 1, 2, \dots, p - 1}}$ [2010]

Q. The number of $A$ in $T_{p}$ such that $\det (A)$ is not divisible by $p$ is

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	Column I		Column II
(A)	$L_{1}, L_{2}, L_{3}$ are concurrent, if	(p)	$k = - 9$
(B)	One of $L_{1}, L_{2}, L_{3}$ is parallel to at least one of the other two, if	(q)	$k = - \frac{6}{5}$
(C)	$L_{1}, L_{2}, L_{3}$ form a triangle, if	(r)	$k = \frac{5}{6}$
(D)	$L_{1}, L_{2}, L_{3}$ do not form a triangle, if	(s)	$k = 5$

Topic Question Set

Q 1 : Let ω≠1 be a cube root of unity and S be the set of all non-singular matrices of the form [1abω1cω2ω1] where each of a, b and c is either ω or ω2. Then the number of distinct matrices in the set S is [2011]

Q 2 : Consider three points P=(-sin(β-α),-cosβ), Q=(cos(β-α),sinβ) and R=(cos(β-α+θ),sin(β-θ)), where 0<α,β,θ<π4. Then, [2008]

Q 3 : Let S={A=(01c1ad1be): a,b,c,d,e∈{0,1} and |A|∈{-1,1}}, where |A| denotes the determinant of A. Then the number of elements in S is ______________. [2024]

Q 4 : The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2×2 matrix such that the trace of A is 3 and the trace of A3 is -18, then the value of the determinant of A is _____. [2020]

Q 5 : Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

Q 6 : Let z=-1+3i2, where i=-1, and r,s∈{1,2,3}. Let P=[(-z)rz2sz2szr] and I be the identity matrix of order 2. Then the total number of ordered pairs (r,s) for which P2=-I is [2016]

Q 7 : Let I=(1001) and P=(2003). Let Q=(xyz4) for some non-zero real numbers x, y and z, for which there is a 2×2 matrix R with all entries being non-zero real numbers, such that QR=RP. Then which of the following statement(s) is (are) TRUE? [2025]

Q 8 : Let M and N be two 3×3 matrices such that MN=NM. Further, if M≠N2 and M2=N4, then [2014]

Q 10 : Let p be an odd prime number and Tp be the following set of 2×2 matrices: Tp={A=[abca]: a,b,c∈{0,1,2,…,p-1}} [2010] Q. The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det(A) is divisible by p, is

Q 11 : Let p be an odd prime number and Tp be the following set of 2×2 matrices: Tp={A=[abca]: a,b,c∈{0,1,2,…,p-1}} [2010] Q. The number of A in Tp 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.]

Q 12 : Let p be an odd prime number and Tp be the following set of 2×2 matrices: Tp={A=[abca]: a,b,c∈{0,1,2,…,p-1}} [2010] Q. The number of A in Tp such that det(A) is not divisible by p is

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Q 2 :

Consider three points

$P = (- \sin (β - α), - \cos β), Q = (\cos (β - α), \sin β)$ and

$R = (\cos (β - α + θ), \sin (β - θ)),$ where $0 < α, β, θ < \frac{π}{4} .$

Then, [2008]

Q 3 :

Let $S = {A = (\begin{matrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{matrix}) : a, b, c, d, e \in {0, 1} and | A | \in {- 1, 1}},$ where $| A |$ denotes the determinant of $A$ .

Then the number of elements in $S$ is ______________. [2024]

Q 4 :

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is 3 and the trace of $A^{3}$ is $- 18$ , then the value of the determinant of $A$ is _____. [2020]

Q 5 :

Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is _____. [2018]

Q 8 :

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N = N M .$ Further, if $M \neq N^{2}$ and $M^{2} = N^{4},$ then [2014]

Q 12 :

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

$T_{p} = {A = [\begin{matrix} a & b \\ c & a \end{matrix}] : a, b, c \in {0, 1, 2, \dots, p - 1}}$ [2010]

Q. The number of $A$ in $T_{p}$ such that $\det (A)$ is not divisible by $p$ is