Matrices and Determinants jee mains questions | JEE Main Maths Matrices and Determinants Previous Year Question

Q 1 :

Consider the matrix $f (x) =$ $[\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$ .

Given below are two statements:

Statement I : $f (- x)$ is the inverse of the matrix $f (x) .$

Statement II : $f (x) f (y) = f (x + y) .$

In the light of the above statements, choose the correct answer from the options given below [2024]

Statement I is true but Statement II is false
Both Statement I and Statement II are false
Statement I is false but Statement II is true
Both Statement I and Statement II are true

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4

(4)

We have, $f (x) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$

$f (- x) = [\begin{matrix} \cos x & \sin x & 0 \\ - \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$

$∴$ $f (x) f (- x) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} \cos x & \sin x & 0 \\ - \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I \Rightarrow f (- x) = {[f (x)]}^{- 1}$

$∴$ Statement-I is true.

$f (x) f (y) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} \cos y & - \sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{matrix}]$

$= [\begin{matrix} \cos x \cos y - \sin x \sin y & \cos x (- \sin y) - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & \sin x (- \sin y) + \cos x \cos y & 0 \\ 0 & 0 & 1 \end{matrix}]$

$= [\begin{matrix} \cos (x + y) & - \sin (x + y) & 0 \\ \sin (x + y) & \cos (x + y) & 0 \\ 0 & 0 & 1 \end{matrix}] = f (x + y)$

$∴$ Statement-II is also true.

Q 2 :

Let $A$ be a square matrix such that $A A^{T} = I .$ Then $\frac{1}{2} A [{(A + A^{T})}^{2} + {(A - A^{T})}^{2}]$ is equal to [2024]

$A^{2} + A^{T}$
$A^{2} + I$
$A^{3} + I$
$A^{3} + A^{T}$

1

2

3

4

(4)

We have, $\frac{1}{2} A [{(A + A^{T})}^{2} + {(A - A^{T})}^{2}]$

$= \frac{1}{2} A [(A^{2} + {(A^{T})}^{2}) + 2 A A^{T} + A^{2} + {(A^{T})}^{2} - 2 A A^{T}]$

$= \frac{1}{2} A [2 A^{2} + 2 {(A^{T})}^{2}] = \frac{1}{2} \times 2 A [A^{2} + {(A^{T})}^{2}]$

$= A^{3} + A A^{T} \cdot A^{T}$

$= A^{3} + I \cdot A^{T}$ [Given, $A A^{T} = I$ ]

$= A^{3} + A^{T}$

Q 3 :

Let $A$ be a $3 \times 3$ real matrix such that $A (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) = 2 (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), A (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) = 4 (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}), A (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) = 2 (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) .$

Then, the system $(A - 3 I)$ $(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix})$ has [2024]

exactly two solutions
unique solution
infinitely many solutions
no solution

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3

4

(2)

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given, $A [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 2 \end{matrix}] \Rightarrow [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}] [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 2 \end{matrix}] \Rightarrow \begin{matrix} x_{1} + z_{1} = 2 \\ x_{2} + z_{2} = 0 \\ x_{3} + z_{3} = 2 \end{matrix}}$ ...(i)

$A [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} - 4 \\ 0 \\ 4 \end{matrix}] \Rightarrow [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}] [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} - 4 \\ 0 \\ 4 \end{matrix}] \Rightarrow \begin{matrix} - x_{1} + z_{1} = - 4 \\ - x_{2} + z_{2} = 0 \\ - x_{3} + z_{3} = 4 \end{matrix}}$ (ii)

and $A [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}] = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}]$

$\Rightarrow y_{1} = 0, y_{2} = 2, y_{3} = 0$

Solving (i) and (ii), we get $x_{1} = 3, x_{2} = 0, x_{3} = - 1, y_{1} = 0, y_{2} = 2, y_{3} = 0$ and $z_{1} = - 1, z_{2} = 0, z_{3} = 3$

$\Rightarrow A = [\begin{matrix} 3 & 0 & - 1 \\ 0 & 2 & 0 \\ - 1 & 0 & 3 \end{matrix}]$

Now, $[\begin{matrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

$\Rightarrow - z = 3, - y = 2 and - x = 1$

$\Rightarrow x = - 1, y = - 2 and z = - 3$

Hence, the given system has unique solution.

Q 4 :

Let $A = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] .$ If the sum of the diagonal elements of $A^{13}$ is $3^{n}$ , then $n$ is equal to ________ . [2024]

(7)

We have, $A = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}]$

$A^{2} = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] = [\begin{matrix} 3 & - 3 \\ 3 & 0 \end{matrix}]$

$A^{3} = [\begin{matrix} 3 & - 3 \\ 3 & 0 \end{matrix}] [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] = [\begin{matrix} 3 & - 6 \\ 6 & - 3 \end{matrix}]$

$A^{4} = [\begin{matrix} 3 & - 3 \\ 3 & 0 \end{matrix}] [\begin{matrix} 3 & - 3 \\ 3 & 0 \end{matrix}] = [\begin{matrix} 0 & - 9 \\ 9 & - 9 \end{matrix}]$

$A^{6} = [\begin{matrix} 3 & - 3 \\ 3 & 0 \end{matrix}] [\begin{matrix} 0 & - 9 \\ 9 & - 9 \end{matrix}] = [\begin{matrix} - 27 & 0 \\ 0 & - 27 \end{matrix}]$

$A^{6} = - 27 I$

$A^{12} = A^{6} A^{6} = (- 27 I) \times (- 27) I$

$\Rightarrow A^{12} = [\begin{matrix} 729 & 0 \\ 0 & 729 \end{matrix}] = 729 I$

So, $A^{13} = 729 I \cdot A = 3^{6} \cdot A = 3^{6} [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] = [\begin{matrix} 3^{6} \times 2 & 3^{6} (- 1) \\ 3^{6} & 3^{6} \end{matrix}]$

Sum of diagonal elements $= 3^{6} \times 2 + 3^{6} = 3^{6} \times 3 = 3^{7}$

$\Rightarrow n = 7$

Q 5 :

Let $A = I_{2} - 2 M M^{T},$ where $M$ is a real matrix of order $2 \times 1$ such that the relation $M^{T} M = I_{1}$ holds. If $λ$ is a real number such that the relation $A X = λ X$ holds for some non-zero real matrix $X$ of order $2 \times 1,$ then the sum of squares of all possible values of $λ$ is equal to _______ . [2024]

(2)

Let $M = [\begin{matrix} a \\ b \end{matrix}]$

Then $M^{T} M = I_{1} \Rightarrow [\begin{matrix} a & b \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = [1]$

$\Rightarrow a^{2} + b^{2} = 1$ ...(i)

Now, $A = - 2 M M^{T} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - 2 [\begin{matrix} a^{2} & a b \\ a b & b^{2} \end{matrix}]$

$\Rightarrow A = [\begin{matrix} 1 - 2 a^{2} & - 2 a b \\ - 2 a b & 1 - 2 b^{2} \end{matrix}]$

Also, Let $X = [\begin{matrix} x \\ y \end{matrix}]$

Then, $A X = λ X$

$\Rightarrow [\begin{matrix} 1 - 2 a^{2} & - 2 a b \\ - 2 a b & 1 - 2 b^{2} \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} λ x \\ λ y \end{matrix}]$

$\Rightarrow [\begin{matrix} (1 - 2 a^{2}) x - 2 a b y \\ - 2 a b x + (1 - 2 b^{2}) y \end{matrix}] = [\begin{matrix} λ x \\ λ y \end{matrix}]$

On comparing, we get

$(1 - 2 a^{2}) x - 2 a b y = λ x$ ...(ii)

$- 2 a b x + (1 - 2 b^{2}) y = λ y$ ...(iii)

From (ii) & (iii), we get

$(1 - 2 a^{2} - λ) (1 - 2 b^{2} - λ) = 4 a^{2} b^{2}$

$\Rightarrow 1 - 2 a^{2} - λ - 2 b^{2} + 4 a^{2} b^{2} + 2 b^{2} λ - λ + 2 a^{2} λ + λ^{2} = 4 a^{2} b^{2}$

$\Rightarrow λ^{2} - 2 λ (1 - a^{2} - b^{2}) + (1 - 2 a^{2} - 2 b^{2}) = 0$

$\Rightarrow λ = \frac{2 (1 - a^{2} - b^{2}) \pm \sqrt{4 {(1 - a^{2} - b^{2})}^{2} - 4 (1 - 2 a^{2} - 2 b^{2})}}{2}$

$\Rightarrow λ = \frac{2 (1 - a^{2} - b^{2}) \pm 2 (a^{2} + b^{2})}{2}$

$\Rightarrow λ = 1 - a^{2} - b^{2} + a^{2} + b^{2} or λ = 1 - a^{2} - b^{2} - a^{2} - b^{2}$

$\Rightarrow λ = 1 or λ = 1 - 2 (a^{2} + b^{2}) = 1 - 2 = - 1$ ( $∵$ from (i))

So, sum of squares of all possible values of $λ = {(1)}^{2} + {(- 1)}^{2} = 2$

Q 6 :

Let A be a $3 \times 3$ real matrix such that $A^{2} (A - 2 I) - 4 (A - I) = O$ , where $I$ and O are the identity and null matrices, respectively. If $A^{5} = α A^{2} + β A + γ I$ , where $α, β$ and $γ$ are real constants, then $α + β + γ$ is equal to : [2025]

20
76
12
4

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3

4

(3)

We have, $A^{2} (A - 2 I) - 4 (A - I) = O$

$\Rightarrow A^{3} - 2 A^{2} - 4 A + 4 I = O$

$\Rightarrow A^{3} = 2 A^{2} + 4 A - 4 I$ ... (i)

Now, $A^{4} = A \cdot A^{3} = A (2 A^{2} + 4 A - 4 I)$

$= 2 A^{3} + 4 A^{2} - 4 A = 2 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 4 A$ [From (i)]

$= 4 A^{2} + 8 A - 8 I + 4 A^{2} - 4 A = 8 A^{2} + 4 A - 8 I$

Similarly, $A^{5} = A \cdot A^{4} = A (8 A^{2} + 4 A - 8 I)$

$= 8 A^{3} + 4 A^{2} - 8 A = 8 (2 A^{2} + 4 A - 4 I) + 4 A^{2} - 8 A$ [From (i)]

$= 20 A^{2} + 24 A - 32 I$

$\Rightarrow α = 20, β = 24 and γ = - 32$

$∴ α + β + γ = 20 + 24 - 32 = 12$ .

Q 7 :

Let the matrix $A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$ satisfy $A^{n} = A^{n - 2} + A^{2} - I$ for $n \geq 3$ . Then the sum of all the elements of $A^{50}$ is: [2025]

44
39
52
53

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(4)

We have, $A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$

$\Rightarrow A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}]$

Since, $A^{n} = A^{n - 2} + A^{2} - I, n > 3$

$\Rightarrow A^{50} = A^{48} + A^{2} - I$

$= [A^{46} + A^{2} - I] + (A^{2} - I)$

$= A^{46} + 2 (A^{2} - I)$

$= A^{44} + 3 (A^{2} - I)$

$= A^{2} + 24 (A^{2} - I)$

$= 25 A^{2} - 24 I$

$= 25 [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] - 24 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{matrix}]$

So, sum of elements of $A^{50} = 53$ .

Q 8 :

Let $α$ be a solution of $x^{2} + x + 1 = 0$ , and for some a and b in R, $[\begin{matrix} 4 & a & b \end{matrix}] [\begin{matrix} 1 & 16 & 13 \\ - 1 & - 1 & 2 \\ - 2 & - 14 & - 8 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \end{matrix}]$ . If $\frac{4}{α^{4}} + \frac{m}{α^{a}} + \frac{n}{α^{b}} = 3$ , then m + n is equal to __________ [2025]

8
3
7
11

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4

(4)

Given, $α$ is a solution of $x^{2} + x + 1 = 0$

$\Rightarrow α^{2} + α + 1 = 0$

$\Rightarrow α = ω or ω^{2}$ (cube root of unity)

Now, $[\begin{matrix} 4 & a & b \end{matrix}] [\begin{matrix} 1 & 16 & 13 \\ - 1 & - 1 & 2 \\ - 2 & - 14 & - 8 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} 4 - a - 2 b & 64 - a - 14 b & 52 + 2 a - 8 b \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \end{matrix}]$

$\Rightarrow a + 2 b = 4; a + 14 b = 64 and a - 4 b = - 26$

On solving above equations we get b = 5 and a = –6

Now, $\frac{4}{α^{4}} + \frac{m}{α^{a}} + \frac{n}{α^{b}} = 3 \Rightarrow \frac{4}{ω^{4}} + \frac{m}{ω^{- 6}} + \frac{n}{ω^{5}} = 3$

$\Rightarrow \frac{4}{ω} + \frac{m}{1} + \frac{n}{ω^{2}} = 3$ [ $∵ ω^{3} = 1$ ]

$\Rightarrow 4 ω^{2} + m + n ω = 3$

$\Rightarrow 4 [\frac{- 1 - \sqrt{3} i}{2}] + m + n [\frac{- 1 + \sqrt{3} i}{2}] = 3$

$\Rightarrow - 2 - 2 \sqrt{3} i + m - \frac{n}{2} + \frac{n \sqrt{3} i}{2} = 3$

$\Rightarrow - 2 + m - \frac{n}{2} = 3 and \frac{n \sqrt{3}}{2} - 2 \sqrt{3} = 0$ [Comparing real and imaginary part]

$\Rightarrow$ m = 7, n = 4 $∴$ m + n = 11.

Q 9 :

Let $A = [a_{i j}]$ be $3 \times 3$ matrix such that $A [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \end{matrix}], A [\begin{matrix} \begin{matrix} 4 \\ 1 \\ 3 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \end{matrix}] and A [\begin{matrix} \begin{matrix} 2 \\ 1 \\ 2 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \end{matrix}]$ , then $a_{23}$ equals: [2025]

–1
0
2
1

1

2

3

4

(1)

Let $A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]$

$A [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \end{matrix}] \Rightarrow [\begin{matrix} \begin{matrix} a_{12} \\ a_{22} \\ a_{33} \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \end{matrix}] \Rightarrow a_{12} = 0; a_{22} = 0; a_{32} = 1$

$A [\begin{matrix} \begin{matrix} 4 \\ 1 \\ 3 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] \end{matrix} \Rightarrow \begin{matrix} 4 a_{11} + a_{12} + 3 a_{13} = 0 & . . . (i) \\ 4 a_{21} + a_{22} + 3 a_{23} \Rightarrow 4 a_{21} + 3 a_{23} = 1 & . . . (ii) \\ 4 a_{31} + a_{32} + 3 a_{33} = 0 & . . . (iii) \end{matrix}$

$A [\begin{matrix} \begin{matrix} 2 \\ 1 \\ 2 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] \end{matrix} \Rightarrow \begin{matrix} 2 a_{11} + a_{12} + 2 a_{13} = 1 & . . . (iv) \\ 2 a_{21} + a_{22} + 2 a_{23} = 0 \Rightarrow a_{21} + a_{23} = 0 & . . . (v) \\ 2 a_{31} + a_{32} + 2 a_{33} = 0 & . . . (vi) \end{matrix}$

On solvig equations (ii) and (v), we get

$a_{21} = 1 and a_{23} = - 1$ .

Q 10 :

Let $A = [\begin{matrix} \frac{1}{\sqrt{2}} & - 2 \\ 0 & 1 \end{matrix}] and P = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}],$ $θ > 0 .$ . If $B = P A P^{T}, C = P^{T} B^{10} P$ and the sum of the diagonal elements of C is $\frac{m}{n}$ , where gcd (m, n) = 1, then m + n is : [2025]

127
2049
258
65

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3

4

(4)

We have, $P = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$

$P^{T} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$

$∴ P P^{T} = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$

$= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$ ... (i)

Now, $B = P A P^{T}$

Pre multiply by $P^{T}$ , we get

$P^{T} B = P^{T} P A P^{T} = A P^{T}$ [using (i)]

Now, post multiply by $P^{T}$ , we get

$P^{T} B P = A P^{T} P = A$

Now, $A^{2} = (P^{T} B P) (P^{T} B P) = P^{T} B^{2} P$

Similarly, $A^{10} = P^{T} B^{10} P = C$ [ $∵ C = P^{T} B^{10} P$ ]

Now, $A^{2} = [\begin{matrix} \frac{1}{\sqrt{2}} & - 2 \\ 0 & 1 \end{matrix}] [\begin{matrix} \frac{1}{\sqrt{2}} & - 2 \\ 0 & 1 \end{matrix}] = [\begin{matrix} {(\frac{1}{\sqrt{2}})}^{2} & - \sqrt{2} - 2 \\ 0 & 1 \end{matrix}]$

Similarly for $A^{3}, A^{4}, . . ., A^{10} = C$

So, sum of diagonal elements of $C = {(\frac{1}{\sqrt{2}})}^{10} + 1$

$= \frac{1}{32} + 1 = \frac{33}{32} = \frac{m}{n}$

$g c d (m, n) \Rightarrow g c d (33, 32) = 1$

So, m + n = 33 + 32 = 65.

Q 11 :

Let $A = [a_{i j}]$ be a matrix of order $3 \times 3$ , with $a_{i j} = {(\sqrt{2})}^{i + j}$ . If the sum of all the elements in the third row of $A^{2}$ is $α + β \sqrt{2}, α, β \in Z$ , then $α + β$ is equal to : [2025]

168
224
210
280

1

2

3

4

(2)

We have,

$A = [a_{i j}] of order 3 \times 3 with a_{i j} = {(\sqrt{2})}^{i + j}$

$A = [\begin{matrix} {(\sqrt{2})}^{2} & {(\sqrt{2})}^{3} & {(\sqrt{2})}^{4} \\ {(\sqrt{2})}^{3} & {(\sqrt{2})}^{4} & {(\sqrt{2})}^{5} \\ {(\sqrt{2})}^{4} & {(\sqrt{2})}^{5} & {(\sqrt{2})}^{6} \end{matrix}] = [\begin{matrix} 2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8 \end{matrix}] = 2 [\begin{matrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{matrix}]$

$\Rightarrow A^{2} = 4 [\begin{matrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{matrix}] [\begin{matrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{matrix}] = 4 [\begin{matrix} 7 & 7 \sqrt{2} & 14 \\ 7 \sqrt{2} & 14 & 14 \sqrt{2} \\ 14 & 14 \sqrt{2} & 28 \end{matrix}]$

$∴$ Sum of elements of third row = $4 (14 + 14 \sqrt{2} + 28)$

$= 4 (42 + 14 \sqrt{2}) = 168 + 56 \sqrt{2}$

Comparing the equation $α + β \sqrt{2}$ , we get

$∴ α = 168, β = 56 \Rightarrow α + β = 224$ .

Q 12 :

Let $A = [\begin{matrix} \cos θ & 0 & - \sin θ \\ 0 & 1 & 0 \\ \sin θ & 0 & \cos θ \end{matrix}]$ . If for some $θ \in (0, π), A^{2} = A^{T}$ , then the sum of the diagonal elements of the matrix ${(A + I)}^{3} + {(A - I)}^{3} - 6 A$ is equal to __________. [2025]

(6)

We have, $A = [\begin{matrix} \cos θ & 0 & - \sin θ \\ 0 & 1 & 0 \\ \sin θ & 0 & \cos θ \end{matrix}]$

Since, A is orthogonal.

$\Rightarrow A A^{T} = A^{T} A = I and A^{T} = A^{- 1}$

Given, $A^{2} = A^{T}$

$\Rightarrow A^{3} = I$

Let $B = {(A + I)}^{3} + {(A - I)}^{3} - 6 A$

$= 2 (A^{3} + 3 A) - 6 A = 2 A^{3} = 2 I$

So, sum of diagonal elements of B = 2(1 + 1 + 1) = 6.

Q 13 :

Let M denote the set of all real matrices of order $3 \times 3$ and let S = {–3, –2, –1, 1, 2}. Let

$S_{1} = {A = [a_{i j}] \in M : A = A^{T} and a_{i j} \in S, \forall i, j}$ ,

$S_{2} = {A = [a_{i j}] \in M : A = - A^{T} and a_{i j} \in S, \forall i, j}$ ,

$S_{3} = {A = [a_{i j}] \in M : a_{11} + a_{22} + a_{33} = 0 and a_{i j} \in S, \forall i, j}$ .

If $n (S_{1} \cup S_{2} \cup S_{3}) = 125 α$ , then $α$ equals __________. [2025]

(1613)

Let M denotes the set of all real matrices of order $3 \times 3$ .

$M = {[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]; a_{i j \in R}}$

Now, $S_{1} = {A = [a_{i j}] \in M : A = A^{T} and a_{i j} \in S, \forall i, j}$

$∴$ Number of elements in $S_{1} = 5^{6}$

$S_{2} = {A = [a_{i j}] \in M : A = - A^{T} and a_{i j} \in S, \forall i, j}$

$\Rightarrow$ Number of elements in $S_{2} = 0$

$S_{3} = {A = [a_{i j}] \in M : a_{11} + a_{22} + a_{33} = 0 and a_{i j} \in S, \forall i, j}$

$\Rightarrow$ Number of elements in $S_{3} = 12 \times 5^{6}$

[ $∵$ Possible cases are (1, 2, –3) $\to$ 3!, (1, 1, –2) $\to$ 3 and (–1, –1, 2) $\to$ 3]

$∴ n (S_{1} \cap S_{3}) = 12 \times 5^{3}$

Now, $n (S_{1} \cup S_{2} \cup S_{3}) = 5^{6} + 12 \times 5^{6} + 0 - 12 \times 5^{3}$

$= 5^{6} (1 + 12) - 12 \times 5^{3} = 5^{3} [13 \times 5^{3} - 12) = 125 α$

$\Rightarrow α = 1613$ .

Q 14 :

Let $S = {m \in Z : A^{m^{2}} + A^{m} = 3 I - A^{- 6}}$ , where $A = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}]$ . Then n(S) is equal to __________. [2025]

(2)

$A = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}], A^{2} = [\begin{matrix} 3 & - 2 \\ 2 & - 1 \end{matrix}], A^{3} = [\begin{matrix} 4 & - 3 \\ 3 & - 2 \end{matrix}]$ ,

$A^{4} = [\begin{matrix} 5 & - 4 \\ 4 & - 3 \end{matrix}]$ and so on $A^{6} = [\begin{matrix} 7 & - 6 \\ 6 & - 5 \end{matrix}]$

$\Rightarrow A^{m} = [\begin{matrix} m + 1 & - m \\ m & - m + 1 \end{matrix}] and A^{m^{2}} = [\begin{matrix} m^{2} + 1 & - m^{2} \\ m^{2} & - (m^{2} - 1) \end{matrix}]$

Now, $A^{m^{2}} + A^{m} = 3 I - A^{- 6}$

$\Rightarrow [\begin{matrix} m^{2} + 1 & - m^{2} \\ m^{2} & - (m^{2} - 1) \end{matrix}] + [\begin{matrix} m + 1 & - m \\ m & - (m - 1) \end{matrix}]$

$= 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} - 5 & 6 \\ - 6 & 7 \end{matrix}] = [\begin{matrix} 8 & - 6 \\ 6 & - 4 \end{matrix}]$

$\Rightarrow m^{2} + 1 + m + 1 \Rightarrow 8 = m^{2} + m - 6 = 0 \Rightarrow m = - 3, 2$

$∴ n (S) = 2$ .

Q 15 :

Let $A = {[a_{i j}]}_{2 \times 2}$ , where $a_{i j} = 0$ for all $i, j$ and $A^{2} = I$ . Let $a$ be the sum of all diagonal elements of $A$ and $b =$ $| A |$ . Then $3 a^{2} + 4 b^{2}$ is equal to [2023]

3
7
4
14

1

2

3

4

(3)

Let
$A = [\begin{matrix} m & n \\ q & p \end{matrix}]$ $∵ A^{2} = I$

$\Rightarrow [\begin{matrix} m^{2} + q n & m n + n p \\ q m + q p & n q + p^{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$\Rightarrow m^{2} + q n = 1$ ...(i); $n (m + p) = 0$ ...(ii)

$q (m + p) = 0$ ...(iii), $n q + p^{2} = 1$ ...(iv)

$\Rightarrow m^{2} - p^{2} = 0 \Rightarrow m = \pm p [Using (i) & (iv)]$

Also, $m + p = 0$

Now, let $a = m + p$ and $b = m p - q n$

$So, 3 a^{2} + 4 b^{2} = 3 \times q^{2} + 4 {(m p - q n)}^{2}$

$= 4 {(- m^{2} - q n)}^{2} = 4 \times 1 = 4$ $[Using (i)]$

Q 16 :

Let $P$ be a square matrix such that $P^{2} = I - P$ . For $α, β, γ, δ \in ℕ$ , if $P^{α} + P^{β} = γ I - 29 P$ and $P^{α} - P^{β} = δ I - 13 P,$ then $α + β + γ - δ$ is equal to [2023]

40
22
18
24

1

2

3

4

(4)

We have,

$p^{α} + p^{β} = γ I - 29 p ...(i)$

$p^{α} - p^{β} = δ I - 13 p ...(ii)$

Now, $p^{2} = I - p$

$p^{3} = p - p^{2} = p - I + p = 2 p - I$

$p^{4} = 2 p^{2} - p = 2 (I - p) - p = 2 I - 3 p$

$p^{5} = 2 p - 3 p^{2} = 2 p - 3 (I - p) = 5 p - 3 I$

$p^{6} = 5 p^{2} - 3 p = 5 (I - p) - 3 p = 5 I - 8 p$

$p^{7} = 5 p - 8 p^{2} = 5 p - 8 (I - p) = 13 p - 8 I$

$p^{8} = 13 p^{2} - 8 p = 13 (I - p) - 8 p = 13 I - 21 p$

$p^{8} + p^{6} = 18 I - 29 p ...(iv)$

$p^{8} - p^{6} = 8 I - 13 p ...(v)$

After comparing (iv) with (i) and (v) with (ii), we get:

$α = 8, β = 6, γ = 18, δ = 8$

$∴ α + β + γ - δ = 8 + 6 + 18 - 8 = 24$

Q 17 :

Let $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}], A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] and Q = P A P^{T} .$ If $P^{T} Q^{2007} P = [\begin{matrix} a & b \\ c & d \end{matrix}],$ then $2 a + b - 3 c - 4 d$ is equal to [2023]

2006
2004
2005
2007

1

2

3

4

(3)

We have, $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}], A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$Q = P A P^{T}$

Also, $P^{T} P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{- 1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}] [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$Q^{2007} = (P A P^{T}) (P A P^{T}) \dots (2007 times) = P A^{2007} P^{T} [∵ P^{T} P = I]$

$∴ P^{T} Q^{2007} P = P^{T} P A^{2007} P^{T} P = A^{2007}$

Now, $A^{2} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$A^{3} = A \cdot A^{2} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}]$

Continuing in the same way, we get:

$A^{2007} = [\begin{matrix} 1 & 2007 \\ 0 & 1 \end{matrix}]$

$∴ P^{T} Q^{2007} P = [\begin{matrix} 1 & 2007 \\ 0 & 1 \end{matrix}] = [\begin{matrix} a & b \\ c & d \end{matrix}] (Given)$

$\Rightarrow a = 1, b = 2007, c = 0, d = 1$

$∴ 2 a + b - 3 c - 4 d = 2 (1) + 2007 - 3 (0) - 4 (1) = 2005$

Q 18 :

Let $A = [\begin{matrix} 1 & \frac{1}{51} \\ 0 & 1 \end{matrix}] .$ If $B = [\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}] A [\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix}],$ then the sum of all the elements of the matrix $\sum_{n = 1}^{50} B^{n}$ is equal to [2023]

50
100
75
125

1

2

3

4

(2)

$A = [\begin{matrix} 1 & \frac{1}{51} \\ 0 & 1 \end{matrix}]; B = [\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}] A [\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

Let $P = [\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}], Q = [\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$B = P A Q$

$B^{2} = (P A Q) (P A Q) = P A Q P A Q = P A^{2} Q$

As, $Q P = [\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$A^{2} = [\begin{matrix} 1 & \frac{1}{51} \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & \frac{1}{51} \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & \frac{2}{51} \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$A^{3} = [\begin{matrix} 1 & \frac{2}{51} \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & \frac{1}{51} \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & \frac{3}{51} \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

Similarly, $A^{n} = [\begin{matrix} 1 & \frac{n}{51} \\ 0 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$B^{n} = P A^{n} Q = [\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}] [\begin{matrix} 1 & \frac{n}{51} \\ 0 & 1 \end{matrix}] [\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix}] \begin{matrix} \end{matrix}$

$B^{n} = [\begin{matrix} 1 + \frac{n}{51} & \frac{n}{51} \\ \frac{- n}{51} & 1 - \frac{n}{51} \end{matrix}] \begin{matrix} \end{matrix}$

$\sum_{n = 1}^{50} B^{n} = [\begin{matrix} 50 + 25 & 25 \\ - 25 & 50 - 25 \end{matrix}] = [\begin{matrix} 75 & 25 \\ - 25 & 25 \end{matrix}] \begin{matrix} \end{matrix}$

Sum of the elements = 100

Q 19 :

The number of symmetric matrices of order 3, with all the entries from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is [2023]

$10^{6}$
$10^{9}$
$6^{10}$
$9^{10}$

1

2

3

4

(1)

Let $A = {[\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}]}_{3 \times 3}$ be the matrix

A is a symmetric matrix

$∴$ $A = A'$

$\Rightarrow [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}] = [\begin{matrix} a & d & g \\ b & e & h \\ c & f & i \end{matrix}] \Rightarrow b = d, c = g, f = h$

$∴$ $A = [\begin{matrix} a & b & c \\ b & e & f \\ c & f & i \end{matrix}]$

Each element is chosen from the set {0, 1, 2, 3, ...., 9}

Choice for each element = 10

$∴ Number of symmetric matrices = 10^{6}$

Q 20 :

If $A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}],$ then [2023]

$A^{30} - A^{25} = 2 I$
$A^{30} = A^{25}$
$A^{30} + A^{25} - A = I$
$A^{30} + A^{25} + A = I$

1

2

3

4

(3)

$A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}] \Rightarrow A = [\begin{matrix} \cos 60 ° & \sin 60 ° \\ - \sin 60 ° & \cos 60 ° \end{matrix}] \begin{matrix} \end{matrix}$

Let $A = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}], where α = \frac{π}{3}$

$A^{2} = [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] [\begin{matrix} \cos α & \sin α \\ - \sin α & \cos α \end{matrix}] \begin{matrix} \end{matrix}$

$= [\begin{matrix} \cos 2 α & \sin 2 α \\ - \sin 2 α & \cos 2 α \end{matrix}] \begin{matrix} \end{matrix}$

$A^{30} = [\begin{matrix} \cos 30 α & \sin 30 α \\ - \sin 30 α & \cos 30 α \end{matrix}]; A^{30} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$

$A^{25} = [\begin{matrix} \cos 25 α & \sin 25 α \\ - \sin 25 α & \cos 25 α \end{matrix}] = [\begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{- \sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$

$\Rightarrow A^{25} = A \Rightarrow A^{25} - A = 0$

Q 21 :

If A and B are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B,$ then [2023]

$A^{2} = I$ or $B = I$
$A^{2} B = B A^{2}$
$A B = I$
$A^{2} B = I$

1

2

3

4

(2)

$A^{2} + B = A^{2} B \Rightarrow A^{2} + B - A^{2} B = 0$

$\Rightarrow A^{2} - A^{2} B + B = 0 \Rightarrow A^{2} (I - B) + B = 0$

$\Rightarrow A^{2} (I - B) + B - I = - I \Rightarrow (A^{2} - I) (I - B) = - I$

$\Rightarrow (A^{2} - I) (B - I) = I \Rightarrow A^{2} B - A^{2} - B = 0$

$\Rightarrow A^{2} (B - I) = B \Rightarrow A^{2} = B {(B - I)}^{- 1} \Rightarrow A^{2} = B (A^{2} - I)$

$\Rightarrow A^{2} = B A^{2} - B \Rightarrow A^{2} + B = B A^{2} \Rightarrow A^{2} B = B A^{2}$

Q 22 :

Let $A = [\begin{matrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{- 3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{matrix}] and B = [\begin{matrix} 1 & - i \\ 0 & 1 \end{matrix}],$ where $i = \sqrt{- 1}$ . If $M = A^{T} B A$ , then the inverse of the matrix $A M^{2023} A^{T}$ is [2023]

$[\begin{matrix} 1 & - 2023 i \\ 0 & 1 \end{matrix}]$
$[\begin{matrix} 1 & 0 \\ 2023 i & 1 \end{matrix}]$
$[\begin{matrix} 1 & 2023 i \\ 0 & 1 \end{matrix}]$
$[\begin{matrix} 1 & 0 \\ - 2023 i & 1 \end{matrix}]$

1

2

3

4

(3)

Q 23 :

Let A, B, C be $3 \times 3$ matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements:

$(S1) A^{13} B^{26} - B^{26} A^{13} is symmetric$

$(S2) A^{26} C^{13} - C^{13} A^{26} is symmetric$

Then, [2023]

Only S1 is true
Both S1 and S2 are false
Both S1 and S2 are true
Only S2 is true

1

2

3

4

(4)

$Since A is symmetric. ∴ A' = A$

$and B and C are skew symmetric.$

$∴ B' = - B and C' = - C$

$S 1 : (A^{13} B^{26} - B^{26} A^{13})' = (A^{13} B^{26})' - (B^{26} A^{13})'$

$= {(B')}^{26} {(A')}^{13} - {(A')}^{13} {(B')}^{26} (∵ (X' Y') = (Y' X'))$

$= {(- B)}^{26} {(A)}^{13} - {(A)}^{13} {(- B)}^{26}$

$= B^{26} A^{13} - A^{13} B^{26} = - (A^{13} B^{26} - B^{26} A^{13})$

Which is skew symmetric.

$∴ S 1 is false.$

$S 2 : (A^{26} C^{13} - C^{13} A^{26})' = (A^{26} C^{13})' - (C^{13} A^{26})'$

$= {(C')}^{13} {(A')}^{26} - {(A')}^{26} {(C')}^{13}$

$= {(- C')}^{13} A^{26} - A^{26} {(- C)}^{13} = - C^{13} A^{26} + A^{26} C^{13}$

Which is symmetric.

$∴ S 2 is true.$

Q 24 :

Let $α$ and $β$ be real numbers. Consider a $3 \times 3$ matrix A such that $A^{2} = 3 A + α I .$ If $A^{4} = 21 A + β I,$ then [2023]

$β = - 8$
$β = 8$
$α = 4$
$α = 1$

1

2

3

4

(1)

$Given, A^{2} = 3 A + α I and A^{4} = 21 A + β I$

$Now, A^{2} A^{2} = (3 A + α I) (3 A + α I)$

$= 9 A^{2} + 6 α I A + α^{2} I = 9 (3 A + α I) + 6 α I A + α^{2} I$

$= 27 A + 9 α I + 6 α A + α^{2} I = (27 + 6 α) A + (9 α + α^{2}) I$

$So, 27 + 6 α = 21 \Rightarrow α = - 1$

$\Rightarrow 9 α + α^{2} = β \Rightarrow 9 (- 1) + {(- 1)}^{2} = β \Rightarrow β = - 8$

Q 25 :

Let $A = (\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}) .$ Then the sum of the diagonal elements of the matrix ${(A + I)}^{11}$ is equal to:

2050
4094
6144
4097

1

2

3

4

(4)

We have, $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] \begin{matrix} \end{matrix}$

$A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] = A$

$Similarly, A^{n} = A; n \in N$

$Now, {(A + I)}^{11} = {}^{11}C_{0} A^{11} + {}^{11}C_{1} A^{10} I^{1} + {}^{11}C_{2} A^{9} I^{2} + \dots + {}^{11}C_{11} A^{0} I^{11}$

$= A ({}^{11}C_{0} + {}^{11}C_{1} + \dots + {}^{11}C_{10}) + I [∵ A^{n} = A and I^{n} = I]$

$= A (2^{11} - 1) + I$

$Trace of {(A + I)}^{11} = 1 (2^{11} - 1) + 1 + 4 (2^{11} - 1) + 1 + (- 3) (2^{11} - 1) + 1$

$= 2^{11} + 4 (2^{11}) - 3 - 3 (2^{11}) + 4 = 2 \times 2^{11} + 1$

$= 2^{12} + 1 = 4096 + 1 = 4097$

Q 26 :

$Let A = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}], where a, c \in ℝ . If A^{3} = A and the positive value of a belongs to the interval$ $(n - 1, n],$

$where n \in ℕ, then n is equal to$ _________ . [2023]

(2)

We have,

$A = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}] and A^{3} = A$

Now, $A^{2} = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}] \begin{matrix} \end{matrix}$

$\Rightarrow A^{2} = [\begin{matrix} a + 2 & 2 c & 3 \\ 3 & a + 3 c & 2 a \\ a c & 1 & 2 + 3 c \end{matrix}] \begin{matrix} \end{matrix}$

and $A^{3} = [\begin{matrix} a + 2 & 2 c & 3 \\ 3 & a + 3 c & 2 a \\ a c & 1 & 2 + 3 c \end{matrix}] [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}]$

$\Rightarrow A^{3} = [\begin{matrix} 2 a c + 3 & a + 2 + 3 c & 2 a + 4 + 6 c \\ a (a + 3 c) + 2 a & 3 + 2 a c & 6 + 3 a + 9 c \\ a + 2 + 3 c & a c + c (2 + 3 c) & 2 a c + 3 \end{matrix}]$

Given that $A^{3} = A$ ,

$2 a c + 3 = 0 ...(i)$

and $a + 2 + 3 c = 1 \Rightarrow a + 1 + 3 c = 0 ...(ii)$

$\Rightarrow a + 1 - \frac{9}{2 a} = 0$ [From (i)]

$\Rightarrow 2 a^{2} + 2 a - 9 = 0$

Since $f (1) < 0, f (2) > 0$

$∴ a \in (1, 2]$ $[∵ f (a) = 2 a^{2} + 2 a - 9]$

Now, $n - 1 = 1 \Rightarrow n = 2$

Q 27 :

$Let A be a symmetric matrix such that | A | = 2 and$ $[\begin{matrix} 2 & 1 \\ 3 & \frac{3}{2} \end{matrix}]$ $A = [\begin{matrix} 1 & 2 \\ α & β \end{matrix}] .$ If the sum of the diagonal elements of A is $s$ , then $\frac{β s}{α^{2}}$ is equal to ________ . [2023]

(5)

Given, $| A | = 2$

Let $A = [\begin{matrix} a & b \\ b & c \end{matrix}] \Rightarrow a c - b^{2} = 2 ...(i)$

Also, $[\begin{matrix} 2 & 1 \\ 3 & \frac{3}{2} \end{matrix}] [\begin{matrix} a & b \\ b & c \end{matrix}] = [\begin{matrix} 1 & 2 \\ α & β \end{matrix}]$

$\Rightarrow [\begin{matrix} 2 a + b & 2 b + c \\ 3 a + \frac{3}{2} b & 3 b + \frac{3}{2} c \end{matrix}] = [\begin{matrix} 1 & 2 \\ α & β \end{matrix}]$

$\Rightarrow 2 a + b = 1 ...(ii), 2 b + c = 2 ...(iii)$

Solving (i), (ii), and (iii) we get $b = - \frac{1}{2}, a = \frac{3}{4}, c = 3$

Now, $α = 3 a + \frac{3}{2} b = \frac{3}{2}, β = 3 b + \frac{3}{2} c = 3$

$s = a + c = \frac{15}{4}$ $∴$ $\frac{β s}{α^{2}} = \frac{3 \times \frac{15}{4}}{\frac{9}{4}} = 5$

Q 28 :

Let $A = [a_{i j}], a_{i j} \in Z \cap [0, 4], 1 \leq i, j \leq 2$ . The number of matrices A such that the sum of all entries is a prime number $p \in (2, 13)$ is ________ . [2023]

(204)

As given $a + b + c + d = 3 or 5 or 7 or 11$

If sum = 3, then

Coefficient of $x^{3}$ in ${(1 + x + x^{2} + \dots + x^{4})}^{4}$ is ${(1 - x^{5})}^{4} {(1 - x)}^{- 4}$

$∴$ ${}^{4 + 3 - 1}{C_{3} = {}^{6}{C_{3}}} = 20$

If sum = 5, coefficient of $x^{5}$ is $(1 - 4 x^{5}) {(1 - x)}^{- 4}$

$∴$ ${}^{4 + 5 - 1}{C_{5}} - 4 = {}^{8}{C_{5}} - 4 = 52$

If sum = 7, then coefficient of $x^{7}$ is

$(1 - 4 x^{5}) {(1 - x)}^{- 4} \Rightarrow {}^{10}C_{7} - 4 \times {}^{5}{C_{2}} = 80$

If sum = 11

$(1 - 4 x^{5} + 6 x^{10}) {(1 - x)}^{- 4}$ $\Rightarrow {}^{4 + 11 - 1}{C_{11}} - 4 {}^{4 + 6 - 1}{C_{6}} + 6 {}^{4 + 1 - 1}{C_{1}}$

$= {}^{14}{C_{11}} - 4$ ${}^{9}{C_{6}} + 24 = 364 - 336 + 24 = 52$

$∴ Total matrices = 20 + 52 + 80 + 52 = 204$

Q 29 :

Let $A = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}]$ and B be two matrices such that $A^{100} = 100 B + I$ . Then the sum of all the elements of $B^{100}$ is ________. [2026]

(0)

$A = I + [\begin{matrix} 2 & - 4 \\ 1 & - 2 \end{matrix}], let M = [\begin{matrix} 2 & - 4 \\ 1 & - 2 \end{matrix}]$

$M^{2} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \Rightarrow M^{3} = M^{4} = \dots = M^{100} = 0$

$A^{100} = {(I + M)}^{100} = \sum_{r = 0}^{100} (\binom{100}{r}) M^{r} . I$

$A^{100} = I + 100 M = I + 100 B$

$∴$ $M = B \Rightarrow M^{100} = B^{100} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

Q 30 :

Let $A = [[\begin{matrix} 0 & 2 & - 3 \\ - 2 & 0 & 1 \\ 3 & - 1 & 0 \end{matrix}]]$ and B be a matrix such that $B (I - A) = I + A$ Then the sum of the diagonal elements of $B^{T} B$ is equal to______. [2026]

(3)

$A^{T} = - A$

$B = (I + A) {(I - A)}^{- 1}$

$B^{T} = ({{(I - A)}^{- 1})}^{T} {(I + A)}^{T}$

$B^{T} = {(I - A^{T})}^{- 1} (I + A^{T})$

$B^{T} = {(I + A)}^{- 1} (I - A)$

$B^{T} B = {(I + A)}^{- 1} (I - A) (I + A) {(I - A)}^{- 1}$

$= {(I + A)}^{- 1} (I + A) (I - A) {(I - A)}^{- 1}$

$= I$

$tr (B^{T} B) = 3$

Topic Question Set

Q 2 :

Let $A$ be a square matrix such that $A A^{T} = I .$ Then $\frac{1}{2} A [{(A + A^{T})}^{2} + {(A - A^{T})}^{2}]$ is equal to [2024]

Q 4 :

Let $A = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] .$ If the sum of the diagonal elements of $A^{13}$ is $3^{n}$ , then $n$ is equal to ________ . [2024]

Q 6 :

Let A be a $3 \times 3$ real matrix such that $A^{2} (A - 2 I) - 4 (A - I) = O$ , where $I$ and O are the identity and null matrices, respectively. If $A^{5} = α A^{2} + β A + γ I$ , where $α, β$ and $γ$ are real constants, then $α + β + γ$ is equal to : [2025]

Q 7 :

Let the matrix $A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$ satisfy $A^{n} = A^{n - 2} + A^{2} - I$ for $n \geq 3$ . Then the sum of all the elements of $A^{50}$ is: [2025]

Q 11 :

Let $A = [a_{i j}]$ be a matrix of order $3 \times 3$ , with $a_{i j} = {(\sqrt{2})}^{i + j}$ . If the sum of all the elements in the third row of $A^{2}$ is $α + β \sqrt{2}, α, β \in Z$ , then $α + β$ is equal to : [2025]

Q 12 :

Let $A = [\begin{matrix} \cos θ & 0 & - \sin θ \\ 0 & 1 & 0 \\ \sin θ & 0 & \cos θ \end{matrix}]$ . If for some $θ \in (0, π), A^{2} = A^{T}$ , then the sum of the diagonal elements of the matrix ${(A + I)}^{3} + {(A - I)}^{3} - 6 A$ is equal to __________. [2025]

Q 14 :

Let $S = {m \in Z : A^{m^{2}} + A^{m} = 3 I - A^{- 6}}$ , where $A = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}]$ . Then n(S) is equal to __________. [2025]

Q 15 :

Let $A = {[a_{i j}]}_{2 \times 2}$ , where $a_{i j} = 0$ for all $i, j$ and $A^{2} = I$ . Let $a$ be the sum of all diagonal elements of $A$ and $b =$ $| A |$ . Then $3 a^{2} + 4 b^{2}$ is equal to [2023]

Q 16 :

Let $P$ be a square matrix such that $P^{2} = I - P$ . For $α, β, γ, δ \in ℕ$ , if $P^{α} + P^{β} = γ I - 29 P$ and $P^{α} - P^{β} = δ I - 13 P,$ then $α + β + γ - δ$ is equal to [2023]

Q 18 :

Let $A = [\begin{matrix} 1 & \frac{1}{51} \\ 0 & 1 \end{matrix}] .$ If $B = [\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}] A [\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix}],$ then the sum of all the elements of the matrix $\sum_{n = 1}^{50} B^{n}$ is equal to [2023]

Q 19 :

The number of symmetric matrices of order 3, with all the entries from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is [2023]

Q 20 :

If $A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}],$ then [2023]

Q 21 :

If A and B are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B,$ then [2023]

Q 22 :

Let $A = [\begin{matrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{- 3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{matrix}] and B = [\begin{matrix} 1 & - i \\ 0 & 1 \end{matrix}],$ where $i = \sqrt{- 1}$ . If $M = A^{T} B A$ , then the inverse of the matrix $A M^{2023} A^{T}$ is [2023]

(3)

Q 23 :

Let A, B, C be $3 \times 3$ matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements:

$(S1) A^{13} B^{26} - B^{26} A^{13} is symmetric$

$(S2) A^{26} C^{13} - C^{13} A^{26} is symmetric$

Then, [2023]

Q 24 :

Let $α$ and $β$ be real numbers. Consider a $3 \times 3$ matrix A such that $A^{2} = 3 A + α I .$ If $A^{4} = 21 A + β I,$ then [2023]

Q 25 :

Let $A = (\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}) .$ Then the sum of the diagonal elements of the matrix ${(A + I)}^{11}$ is equal to:

Q 26 :

$Let A = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}], where a, c \in ℝ . If A^{3} = A and the positive value of a belongs to the interval$ $(n - 1, n],$

$where n \in ℕ, then n is equal to$ _________ . [2023]

Q 27 :

$Let A be a symmetric matrix such that | A | = 2 and$ $[\begin{matrix} 2 & 1 \\ 3 & \frac{3}{2} \end{matrix}]$ $A = [\begin{matrix} 1 & 2 \\ α & β \end{matrix}] .$ If the sum of the diagonal elements of A is $s$ , then $\frac{β s}{α^{2}}$ is equal to ________ . [2023]

Q 28 :

Let $A = [a_{i j}], a_{i j} \in Z \cap [0, 4], 1 \leq i, j \leq 2$ . The number of matrices A such that the sum of all entries is a prime number $p \in (2, 13)$ is ________ . [2023]

Q 29 :

Let $A = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}]$ and B be two matrices such that $A^{100} = 100 B + I$ . Then the sum of all the elements of $B^{100}$ is ________. [2026]

Q 30 :

Let $A = [[\begin{matrix} 0 & 2 & - 3 \\ - 2 & 0 & 1 \\ 3 & - 1 & 0 \end{matrix}]]$ and B be a matrix such that $B (I - A) = I + A$ Then the sum of the diagonal elements of $B^{T} B$ is equal to______. [2026]

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Topic Question Set

Q 1 : Consider the matrix f(x)=[cosx-sinx0sinxcosx0001]. Given below are two statements: Statement I : f(-x) is the inverse of the matrix f(x). Statement II : f(x)f(y)=f(x+y). In the light of the above statements, choose the correct answer from the options given below [2024]

Q 2 : Let A be a square matrix such that AAT=I. Then 12A[(A+AT)2+(A-AT)2] is equal to [2024]

Q 3 : Let A be a 3×3 real matrix such that A(101)=2(100), A(-101)=4(-101), A(010)=2(010). Then, the system (A-3I) (xyz)=(123) has [2024]

Q 4 : Let A=[2-111]. If the sum of the diagonal elements of A13 is 3n, then n is equal to ________ . [2024]

Q 5 : Let A=I2-2MMT, where M is a real matrix of order 2×1 such that the relation MTM=I1 holds. If λ is a real number such that the relation AX=λX holds for some non-zero real matrix X of order 2×1, then the sum of squares of all possible values of λ is equal to _______ . [2024]

Q 6 : Let A be a 3×3 real matrix such that A2(A–2I)–4(A–I)=O, where I and O are the identity and null matrices, respectively. If A5=αA2+βA+γI, where α, β and γ are real constants, then α+β+γ is equal to : [2025]

Q 7 : Let the matrix A=[100101010] satisfy An=An–2+A2–I for n≥3. Then the sum of all the elements of A50 is: [2025]

Q 8 : Let α be a solution of x2+x+1=0, and for some a and b in R, [4ab][11613–1–12–2–14–8][000]. If 4α4+mαa+nαb=3, then m + n is equal to __________ [2025]

Q 9 : Let A=[aij] be 3×3 matrix such that A[010]=[001], A[413]=[010] and A[212]=[100], then a23 equals: [2025]

Q 10 : Let A=[12–201] and P=[cos θ– sin θsin θcos θ],θ>0.. If B=PAPT,C=PTB10P and the sum of the diagonal elements of C is mn, where gcd (m, n) = 1, then m + n is : [2025]

Q 11 : Let A=[aij] be a matrix of order 3×3, with aij=(2)i+j. If the sum of all the elements in the third row of A2 is α+β2, α, β∈Z, then α+β is equal to : [2025]

Q 12 : Let A=[cosθ0–sinθ010sinθ0cosθ]. If for some θ∈(0,π), A2=AT, then the sum of the diagonal elements of the matrix (A+I)3+(A–I)3–6A is equal to __________. [2025]

Q 14 : Let S={m∈Z : Am2+Am=3I–A–6}, where A=[2–110]. Then n(S) is equal to __________. [2025]

Q 15 : Let A=[aij]2×2, where aij=0 for all i,j and A2=I. Let a be the sum of all diagonal elements of A and b=|A|. Then 3a2+4b2 is equal to [2023]

Q 16 : Let P be a square matrix such that P2=I-P. For α,β,γ,δ∈ℕ, if Pα+Pβ=γI-29P and Pα-Pβ=δI-13P, then α+β+γ-δ is equal to [2023]

Q 17 : Let P= [3212-1232], A=[1101] and Q=PAPT. If PTQ2007P=[abcd], then 2a+b-3c-4d is equal to [2023]

Q 18 : Let A=[115101]. If B=[12-1-1]A[-1-211], then the sum of all the elements of the matrix ∑n=150Bn is equal to [2023]

Q 19 : The number of symmetric matrices of order 3, with all the entries from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is [2023]

Q 20 : If A=12[13-31], then [2023]

Q 21 : If A and B are two non-zero n×n matrices such that A2+B=A2B, then [2023]

Q 22 : Let A=[110310-310110] and B=[1-i01], where i=-1. If M=ATBA, then the inverse of the matrix AM2023AT is [2023]

(3)

Q 23 : Let A, B, C be 3×3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements: (S1) A13B26-B26A13 is symmetric (S2) A26C13-C13A26 is symmetric Then, [2023]

Q 24 : Let α and β be real numbers. Consider a 3×3 matrix A such that A2=3A+αI. If A4=21A+βI, then [2023]

Q 25 : Let A=(10004-1012-3). Then the sum of the diagonal elements of the matrix (A+I)11 is equal to:

Q 26 : Let A=[012a031c0], where a,c∈ℝ. If A3=A and the positive value of a belongs to the interval (n-1,n], where n∈ℕ, then n is equal to _________ . [2023]

Q 27 : Let A be a symmetric matrix such that |A|=2 and [21332]A=[12αβ]. If the sum of the diagonal elements of A is s, then βsα2 is equal to ________ . [2023]

Q 28 : Let A=[aij], aij∈Z∩[0,4], 1≤i,j≤2. The number of matrices A such that the sum of all entries is a prime number p∈(2,13) is ________ . [2023]

Q 29 : Let A=[3-41-1] and B be two matrices such that A100=100B+I. Then the sum of all the elements of B100 is ________. [2026]

Q 30 : Let A=[02-3-2013-10] and B be a matrix such that B(I-A)=I+A Then the sum of the diagonal elements of BTB is equal to______. [2026]

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Q 2 :

Let $A$ be a square matrix such that $A A^{T} = I .$ Then $\frac{1}{2} A [{(A + A^{T})}^{2} + {(A - A^{T})}^{2}]$ is equal to [2024]

Q 4 :

Let $A = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}] .$ If the sum of the diagonal elements of $A^{13}$ is $3^{n}$ , then $n$ is equal to ________ . [2024]

Q 6 :

Let A be a $3 \times 3$ real matrix such that $A^{2} (A - 2 I) - 4 (A - I) = O$ , where $I$ and O are the identity and null matrices, respectively. If $A^{5} = α A^{2} + β A + γ I$ , where $α, β$ and $γ$ are real constants, then $α + β + γ$ is equal to : [2025]

Q 7 :

Let the matrix $A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$ satisfy $A^{n} = A^{n - 2} + A^{2} - I$ for $n \geq 3$ . Then the sum of all the elements of $A^{50}$ is: [2025]

Q 11 :

Let $A = [a_{i j}]$ be a matrix of order $3 \times 3$ , with $a_{i j} = {(\sqrt{2})}^{i + j}$ . If the sum of all the elements in the third row of $A^{2}$ is $α + β \sqrt{2}, α, β \in Z$ , then $α + β$ is equal to : [2025]

Q 12 :

Let $A = [\begin{matrix} \cos θ & 0 & - \sin θ \\ 0 & 1 & 0 \\ \sin θ & 0 & \cos θ \end{matrix}]$ . If for some $θ \in (0, π), A^{2} = A^{T}$ , then the sum of the diagonal elements of the matrix ${(A + I)}^{3} + {(A - I)}^{3} - 6 A$ is equal to __________. [2025]

Q 14 :

Let $S = {m \in Z : A^{m^{2}} + A^{m} = 3 I - A^{- 6}}$ , where $A = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}]$ . Then n(S) is equal to __________. [2025]

Q 15 :

Let $A = {[a_{i j}]}_{2 \times 2}$ , where $a_{i j} = 0$ for all $i, j$ and $A^{2} = I$ . Let $a$ be the sum of all diagonal elements of $A$ and $b =$ $| A |$ . Then $3 a^{2} + 4 b^{2}$ is equal to [2023]

Q 16 :

Let $P$ be a square matrix such that $P^{2} = I - P$ . For $α, β, γ, δ \in ℕ$ , if $P^{α} + P^{β} = γ I - 29 P$ and $P^{α} - P^{β} = δ I - 13 P,$ then $α + β + γ - δ$ is equal to [2023]

Q 18 :

Let $A = [\begin{matrix} 1 & \frac{1}{51} \\ 0 & 1 \end{matrix}] .$ If $B = [\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix}] A [\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix}],$ then the sum of all the elements of the matrix $\sum_{n = 1}^{50} B^{n}$ is equal to [2023]

Q 19 :

The number of symmetric matrices of order 3, with all the entries from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is [2023]

Q 20 :

If $A = \frac{1}{2} [\begin{matrix} 1 & \sqrt{3} \\ - \sqrt{3} & 1 \end{matrix}],$ then [2023]

Q 21 :

If A and B are two non-zero $n \times n$ matrices such that $A^{2} + B = A^{2} B,$ then [2023]

Q 22 :

Let $A = [\begin{matrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{- 3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{matrix}] and B = [\begin{matrix} 1 & - i \\ 0 & 1 \end{matrix}],$ where $i = \sqrt{- 1}$ . If $M = A^{T} B A$ , then the inverse of the matrix $A M^{2023} A^{T}$ is [2023]

Q 23 :

Let A, B, C be $3 \times 3$ matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements:

$(S1) A^{13} B^{26} - B^{26} A^{13} is symmetric$

$(S2) A^{26} C^{13} - C^{13} A^{26} is symmetric$

Then, [2023]

Q 24 :

Let $α$ and $β$ be real numbers. Consider a $3 \times 3$ matrix A such that $A^{2} = 3 A + α I .$ If $A^{4} = 21 A + β I,$ then [2023]

Q 25 :

Let $A = (\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}) .$ Then the sum of the diagonal elements of the matrix ${(A + I)}^{11}$ is equal to:

Q 26 :

$Let A = [\begin{matrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{matrix}], where a, c \in ℝ . If A^{3} = A and the positive value of a belongs to the interval$ $(n - 1, n],$

$where n \in ℕ, then n is equal to$ _________ . [2023]

Q 27 :

$Let A be a symmetric matrix such that | A | = 2 and$ $[\begin{matrix} 2 & 1 \\ 3 & \frac{3}{2} \end{matrix}]$ $A = [\begin{matrix} 1 & 2 \\ α & β \end{matrix}] .$ If the sum of the diagonal elements of A is $s$ , then $\frac{β s}{α^{2}}$ is equal to ________ . [2023]

Q 28 :

Let $A = [a_{i j}], a_{i j} \in Z \cap [0, 4], 1 \leq i, j \leq 2$ . The number of matrices A such that the sum of all entries is a prime number $p \in (2, 13)$ is ________ . [2023]

Q 29 :

Let $A = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}]$ and B be two matrices such that $A^{100} = 100 B + I$ . Then the sum of all the elements of $B^{100}$ is ________. [2026]

Q 30 :

Let $A = [[\begin{matrix} 0 & 2 & - 3 \\ - 2 & 0 & 1 \\ 3 & - 1 & 0 \end{matrix}]]$ and B be a matrix such that $B (I - A) = I + A$ Then the sum of the diagonal elements of $B^{T} B$ is equal to______. [2026]