MHT CET · Maths · Matrices
\(A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\) and \(A B=B A=I\), then \(B\) is
equal to
- A \(\left[\begin{array}{rr}-\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\)
- 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 \(\left[\begin{array}{rr}\sin \theta & -\cos \theta \\ -\cos \theta & \sin \theta\end{array}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Given, \(A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\)
and \(\quad A B=B A=I\)
\(\Rightarrow B=A^{-1} I=A^{-1}\)
\(=\frac{1}{\cos ^{2} \theta+\sin ^{2} \theta}\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\)
\(\Rightarrow B=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\)
and \(\quad A B=B A=I\)
\(\Rightarrow B=A^{-1} I=A^{-1}\)
\(=\frac{1}{\cos ^{2} \theta+\sin ^{2} \theta}\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\)
\(\Rightarrow B=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\)
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