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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Matrices

NDA 2016 · Mathematics · Matrices

If m=[ array ll 1 & 0 0 & 1 array ] and n=[ array ll 0 & 1 -1 & 0...

If \mathrm{m}=\left[1001\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] and\mathrm{n}=\left[01−10\begin{array}{ll}0 & 1 \\ -1 & 0\end{array}\right],thenwhatisthevalueofthedeterminantof, then what is the value of the determinant of\mathrm{m} \cos \theta-\mathrm{n} \sin \theta$ ?

Options

  1. A−1-1
  2. B00
  3. C11
  4. D22

Correct answer

C. 11

Step-by-step solution

mcos⁡θ−nsin⁡θ=[cos⁡θ00cos⁡θ]−[0sin⁡θ−sin⁡θ0]=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{aligned} m \cos \theta-n \sin \theta & =\left[\begin{array}{cc}\cos \theta & 0 \\ 0 & \cos \theta\end{array}\right]-\left[\begin{array}{cc}0 & \sin \theta \\ -\sin \theta & 0\end{array}\right] \\ & =\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\end{aligned} det. (mcos⁡θ−nsin⁡θ)=cos⁡2θ−(−sin⁡2θ)=cos⁡2θ+sin⁡2θ=1\begin{aligned} & \text{ det. }(m \cos \theta-n \sin \theta)=\cos ^2 \theta-\left(-\sin ^2 \theta\right) \\ & \quad=\cos ^2 \theta+\sin ^2 \theta=1\end{aligned}$
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