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CBSE Class 12 Math 2024 All Sets Solved Paper

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Question : 17 of 20
Marks: +1, -0
If F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} and [F(x)]2=F(kx)[F(x)]^2 = F(kx), then the value of k is
Solution:  
[F(x)]2=F(kx)[F(x)]^2 = F(kx)
[cos⁡x−sin⁡x0sin⁡xcos⁡x0001][cos⁡x−sin⁡x0sin⁡xcos⁡x0001]=[cos⁡kx−sin⁡kx0sin⁡kxcos⁡kx0001]\begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos kx & -\sin kx & 0 \\ \sin kx & \cos kx & 0 \\ 0 & 0 & 1 \end{bmatrix}
[cos⁡2x−sin⁡2x−cos⁡xsin⁡x−sin⁡xcos⁡x0sin⁡xcos⁡x+cos⁡xsin⁡x−sin⁡2x+cos⁡2x0001]=[cos⁡kx−sin⁡kx0sin⁡kxcos⁡kx0001]\begin{bmatrix} \cos^2 x - \sin^2 x & -\cos x \sin x - \sin x \cos x & 0 \\ \sin x \cos x + \cos x \sin x & -\sin^2 x + \cos^2 x & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos kx & -\sin kx & 0 \\ \sin kx & \cos kx & 0 \\ 0 & 0 & 1 \end{bmatrix}
[cos⁡2x−2sin⁡xcos⁡x02sin⁡xcos⁡xcos⁡2x0001]=[cos⁡kx−sin⁡kx0sin⁡kxcos⁡kx0001]\begin{bmatrix} \cos 2x & -2\sin x \cos x & 0 \\ 2\sin x \cos x & \cos 2x & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos kx & -\sin kx & 0 \\ \sin kx & \cos kx & 0 \\ 0 & 0 & 1 \end{bmatrix}
(by using formula cos⁡2x=cos⁡2x−sin⁡2x)\text{(by using formula } \cos 2x = \cos^2 x - \sin^2 x)
[cos⁡2x−sin⁡2x0sin⁡2xcos⁡2x0001]=[cos⁡kx−sin⁡kx0sin⁡kxcos⁡kx0001]\begin{bmatrix} \cos 2x & -\sin 2x & 0 \\ \sin 2x & \cos 2x & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos kx & -\sin kx & 0 \\ \sin kx & \cos kx & 0 \\ 0 & 0 & 1 \end{bmatrix}
(by using formula sin⁡2x=2sin⁡xcos⁡x)\text{(by using formula } \sin 2x = 2\sin x \cos x)
The above matrix equation, we get cos⁡2x=cos⁡kx\cos 2x = \cos kx and sin⁡2x=sin⁡kx\sin 2x = \sin kx
⇒2x=kx\Rightarrow 2x = kx
⇒k=2\Rightarrow k = 2
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