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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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]

0%

(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]

0%

(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

1

2

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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3

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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2

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

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

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.

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]

0%

0%

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]

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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]

0%

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]

0%

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]

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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]