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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Matrices

NDA 2016 · Mathematics · Matrices

If f\ x\ =[ array ccc x & - x & 0 x & x & 0 0 & 0 & 1 array ] ,...

If f\{x\}=\left[cos⁡x−sin⁡x0sin⁡xcos⁡x0001\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right], then which of the following are correct? 1. f(θ)×f(ϕ)=f(θ+ϕ)2.Thevalueofthedeterminantofthematrix\mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi) 2. The value of the determinant of the matrix f(\theta) \times f(\phi)is1.3.Thedeterminantofis 1. 3. The determinant off(x)isanevenfunction.Selectthecorrectanswerusingthecodegivenbelow:is an even function. Select the correct answer using the code given below:

Options

  1. A1 and 2 only
  2. B2 and 3 only
  3. C1 and 3 only
  4. D1,2 and 3

Correct answer

D. 1,2 and 3

Step-by-step solution

\mathrm{f}(\theta)=\left[cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right] \& \mathrm{f}(\phi)\left[cos⁡ϕ−sin⁡ϕ0sin⁡ϕcos⁡ϕ0001\begin{array}{ccc}\cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1\end{array}\right]f(θ)×f(ϕ)=[cos⁡θcos⁡ϕ−sin⁡θsin⁡ϕ−cos⁡θsin⁡ϕ−sin⁡θcos⁡ϕ0sin⁡θcos⁡ϕ+cos⁡θsin⁡ϕ−sin⁡θsin⁡ϕ+cos⁡θcos⁡ϕ0001] (using Trigonometric Identities) ⇒f(θ)×f(ϕ)=f(θ+ϕ).  Also, det, [f(θ)×f(ϕ)]=1[cos⁡2(θ+ϕ)−(−sin⁡2(θ+ϕ))]=cos⁡2(θ+ϕ)+sin⁡2(θ+ϕ)=1.\begin{aligned} & \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\left[\begin{array}{ccc} \cos \theta \cos \phi-\sin \theta \sin \phi & -\cos \theta \sin \phi-\sin \theta \cos \phi & 0 \\ \sin \theta \cos \phi+\cos \theta \sin \phi & -\sin \theta \sin \phi+\cos \theta \cos \phi & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \text{ (using Trigonometric Identities) } \\ & \Rightarrow \quad \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi) \text{. } \\ & \text{ Also, det, }[\mathrm{f}(\theta) \times \mathrm{f}(\phi)]=1\left[\cos ^2(\theta+\phi)-\left(-\sin ^2(\theta+\phi)\right)\right] \\ & =\cos ^2(\theta+\phi)+\sin ^2(\theta+\phi)=1 . \end{aligned}&det⁡.(f(x))=(cos⁡2x−(−sin⁡2x)=cos⁡2x+sin⁡2x=1\& \operatorname{det} .(\mathrm{f}(x))=\left(\cos ^2 x-\left(-\sin ^2 x\right)=\cos ^2 x+\sin ^2 x=1\right.. For x=−xdet⁡.(f(−x))=cos⁡2(−x)+Sin⁡2(−x)=1x=-x\operatorname{det} .(\mathrm{f}(-x))=\cos ^2(-x)+\operatorname{Sin}^2(-x)=1 Hence, det⁡.(f(−x))=det⁡.(f(x))\operatorname{det} .(\mathrm{f}(-x))=\operatorname{det} .(\mathrm{f}(x)) Hence, det.( f(x)\mathrm{f}(\mathrm{x}) ) is even function.
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