MHT CET · Maths · Matrices
Let \(A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right]\), then the inverse of \(A\)
is
- A \(\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right]\)
- B \(\left[\begin{array}{cc}-\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\)
- C \(\left[\begin{array}{ll}\sin \theta & -\cos \theta \\ \cos \theta & -\sin \theta\end{array}\right]\)
- D \(\left[\begin{array}{cc}-\sin \theta & -\cos \theta \\ -\cos \theta & \sin \theta\end{array}\right]\)
Answer & Solution
Correct Answer
(A) \(\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
\quad|A|=\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
-\sin \theta & -\cos \theta
\end{array}\right|=-1 \\
\operatorname{adj}(A)=\left[\begin{array}{cc}
-\cos \theta & \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right] \\
\therefore \quad A^{-1}=\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
-\sin \theta & -\cos \theta
\end{array}\right|=A
\end{array}
\)
\begin{array}{l}
\quad|A|=\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
-\sin \theta & -\cos \theta
\end{array}\right|=-1 \\
\operatorname{adj}(A)=\left[\begin{array}{cc}
-\cos \theta & \sin \theta \\
\sin \theta & \cos \theta
\end{array}\right] \\
\therefore \quad A^{-1}=\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
-\sin \theta & -\cos \theta
\end{array}\right|=A
\end{array}
\)
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