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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Matrices

NDA 2015 · Mathematics · Matrices

If E(θ)=[ array cc θ & θ - θ & θ array ] , then E(α) E(β) is equal...

If E(\theta)=\left[cos⁡θsin⁡θ−sin⁡θcos⁡θ\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right], then E(α)E(β)E(\alpha) E(\beta) is equal to

Options

  1. AE(αβ)E(\alpha \beta)
  2. BE(α−β)E(\alpha-\beta)
  3. CE(α+β)E(\alpha+\beta)
  4. D−E(α+β)-E(\alpha+\beta)

Correct answer

C. E(α+β)E(\alpha+\beta)

Step-by-step solution

E(θ)=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]Now E(α)=[cos⁡αsin⁡α−sin⁡αcos⁡α]\begin{aligned} & E(\theta)=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] \\ & \text{Now } E(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\end{aligned} and E(β)=[cos⁡βsin⁡β−sin⁡βcos⁡β] Now E(α)E(β)=[cos⁡αsin⁡α−sin⁡αcos⁡α][cos⁡βsin⁡β−sin⁡βcos⁡β]=[cos⁡α⋅cos⁡βcos⁡αsin⁡β−sin⁡αsin⁡β+sin⁡αcos⁡β−sin⁡αcos⁡β−sin⁡αsin⁡β−cos⁡αsin⁡β+cos⁡αcos⁡β]=[cos⁡(α+β)sin⁡(α+β)−sin⁡(α+β)cos⁡(α+β)]=E(α+β)\begin{aligned} & \text{ and } E(\beta)=\left[\begin{array}{cc} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{array}\right] \\ & \text{ Now } E(\alpha) E(\beta)=\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{array}\right] \\ & =\left[\begin{array}{cc} \cos \alpha \cdot \cos \beta & \cos \alpha \sin \beta \\ -\sin \alpha \sin \beta & +\sin \alpha \cos \beta \\ -\sin \alpha \cos \beta & -\sin \alpha \sin \beta \\ -\cos \alpha \sin \beta & +\cos \alpha \cos \beta \end{array}\right] \\ & =\left[\begin{array}{cc} \cos(\alpha+\beta) & \sin(\alpha+\beta) \\ -\sin(\alpha+\beta) & \cos(\alpha+\beta) \end{array}\right] \\ & =E(\alpha+\beta) \end{aligned}\therefore \quad$ Option (C) is correct.
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