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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(D)

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]

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

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(D)

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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(B)

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]

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(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]

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(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 $λ = {(1)}^{2} + {(- 1)}^{2} = 2$

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]

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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]

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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]

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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]