Q59 / 150 · MHT CET 2009 · Mathematics

$A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$ and $A B=B A=I$, then $B$ is equal to

  • A
    $\left[\begin{array}{rr}-\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$
  • Correct answer
    B
    $\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
  • C
    $\left[\begin{array}{rr}-\sin \theta & \cos \theta \\ \cos \theta & \sin \theta\end{array}\right]$
  • D
    $\left[\begin{array}{rr}\sin \theta & -\cos \theta \\ -\cos \theta & \sin \theta\end{array}\right]$

Solution

Given, $A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$ and $\quad A B=B A=I$ $\Rightarrow B=A^{-1} I=A^{-1}$ $=\frac{1}{\cos ^{2} \theta+\sin ^{2} \theta}\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$ $\Rightarrow B=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

Hence, the correct answer is option: 'B'.

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