If n- 2 is an integer A- cos2-n sin2-n 0- -sin2-n cos2-n 0- 0 0 1 - and I is the identity matrix of order 3- Then

Explanation for correct optionOption(A): Given that n≥ 2 is an integer A=[[ cos(2π/n) sin(2π/n) 0; sin(2π/n) cos(2π/n) 0; 0 ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

If n≥ 2 is an...

Question

If $n \geq 2$ is an integer $A = [\begin{matrix} \cos (\frac{2 π}{n}) & \sin (\frac{2 π}{n}) & 0 \\ - \sin (\frac{2 π}{n}) & \cos (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}]$ and $I$ is the identity matrix of order $3$ . Then

$A^{n} = I$ and $A^{n - 1} \neq I$

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$A^{m} \neq I$ for any positive integer $m$

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$A$ is not invertible

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$A^{m} = 0$ for a positive integer $m$

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Solution

The correct option is A
$A^{n} = I$ and $A^{n - 1} \neq I$

Explanation for correct option
Option(A):
Given that $n \geq 2$ is an integer $A = [\begin{matrix} \cos (\frac{2 π}{n}) & \sin (\frac{2 π}{n}) & 0 \\ - \sin (\frac{2 π}{n}) & \cos (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}]$ and $I$ is the identity matrix of order $3$
$∴ A = [\begin{matrix} \cos (\frac{2 π}{n}) & \sin (\frac{2 π}{n}) & 0 \\ - \sin (\frac{2 π}{n}) & \cos (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} \cos (\frac{2 π}{n}) & \sin (\frac{2 π}{n}) & 0 \\ - \sin (\frac{2 π}{n}) & \cos (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}] \times [\begin{matrix} \cos (\frac{2 π}{n}) & \sin (\frac{2 π}{n}) & 0 \\ - \sin (\frac{2 π}{n}) & \cos (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} \cos^{2} (\frac{2 π}{n}) - \sin^{2} (\frac{2 π}{n}) & 2 \cos (\frac{2 π}{n}) \sin (\frac{2 π}{n}) & 0 \\ - 2 \cos (\frac{2 π}{n}) \sin (\frac{2 π}{n}) & \cos^{2} (\frac{2 π}{n}) - \sin^{2} (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} \cos (\frac{4 π}{n}) & \sin (\frac{4 π}{n}) & 0 \\ - \sin (\frac{4 π}{n}) & \cos (\frac{4 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}] [∵ 2 \sin (x) \cos (x) = \sin (2 x), \cos^{2} (x) - \sin^{2} (x) = \cos (2 x)] \Rightarrow A^{2} = [\begin{matrix} \cos (2^{1} (\frac{2 π}{n})) & \sin (2^{1} (\frac{2 π}{n})) & 0 \\ - \sin (2^{1} (\frac{2 π}{n})) & \cos (2^{1} (\frac{2 π}{n})) & 0 \\ 0 & 0 & 1 \end{matrix}]$
Similarly,
$A^{n} = [\begin{matrix} \cos (n (\frac{2 π}{n})) & \sin (n (\frac{2 π}{n})) & 0 \\ - \sin (n (\frac{2 π}{n})) & \cos (n \frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \cos (0) & \sin (0) & 0 \\ - \sin (0) & \cos (0) & 0 \\ 0 & 0 & 1 \end{matrix}] [∵ \cos (0) = 1, \sin (0) = 0] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I ∴ A^{n} = I \Rightarrow A^{- 1} \cdot A^{n} = A^{- 1} \cdot I \Rightarrow A^{n - 1} = A^{- 1} ∵ A \neq I ∴ A^{- 1} \neq I \Rightarrow A^{n - 1} \neq I$
Hence option(A) is correct
Explanation for incorrect options
Option(B), (D): From the above explanation clearly Option(B) and Option(D) are incorrect
Option(C): We can say that a square matrix is invertible if and only if the determinant is not equal to zero
$|A| = |\begin{matrix} \cos (\frac{2 π}{n}) & \sin (\frac{2 π}{n}) & 0 \\ - \sin (\frac{2 π}{n}) & \cos (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}| = \cos (\frac{2 π}{n}) (\cos (\frac{2 π}{n}) - 0) + \sin (\frac{2 π}{n}) (\sin (\frac{2 π}{n}) - 0) + 0 = \cos^{2} (\frac{2 π}{n}) + \sin^{2} (\frac{2 π}{n}) = 1 [∵ \cos^{2} (θ) + \sin^{2} (θ) = 1] \neq 0$
Clearly $A$ is invertible
Hence, option(C) is incorrect
Hence, option(A) is correct

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Similar questions

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If n≥2 is an integer A=cos2πnsin2πn0-sin2πncos2πn0001 and I is the identity matrix of order 3. Then

If $n \geq 2$ is an integer $A = [\begin{matrix} \cos (\frac{2 π}{n}) & \sin (\frac{2 π}{n}) & 0 \\ - \sin (\frac{2 π}{n}) & \cos (\frac{2 π}{n}) & 0 \\ 0 & 0 & 1 \end{matrix}]$ and $I$ is the identity matrix of order $3$ . Then