If $A(\theta)=\left[\begin{array}{cc}i \sin \theta & \cos \theta \\ \cos \theta & i \sin…

If $A(\theta)=\left[\begin{array}{cc}i \sin \theta & \cos \theta \\ \cos \theta & i \sin \theta\end{array}\right]$ is a matrix, where $i=\sqrt{-1}$, then which of the following is not true
  1. $\operatorname{det} A(\pi+\theta)=\operatorname{det} A(-\theta)$
  2. $\operatorname{det} A(-\theta)=\operatorname{det} A(\theta)$
  3. $\operatorname{det}[A(\theta)]^{-1}=1$
  4. $\operatorname{det} A(-\theta)=-1$

Solution

$A(\theta)=\left[\begin{array}{cc}i \sin \theta & \cos \theta \\ \cos \theta & i \sin \theta\end{array}\right]$ where $i=\sqrt{-1}$ $\therefore|A(\theta)|=i^2 \sin ^2 \theta-\cos ^2 \theta$ $=-\left(\sin ^2 \theta+\cos ^2 \theta\right)=-1$ $\operatorname{adj} A(\theta)=\left[\begin{array}{cc}i \sin \theta & -\cos \theta \\ -\cos \theta & i \sin \theta\end{array}\right]$ $\therefore(A(\theta))^{-1}=\frac{1}{-1}\left[\begin{array}{cc}i \sin \theta & -\cos \theta \\ -\cos \theta & i \sin \theta\end{array}\right]$ $\left|A(\theta)^{-1}\right|=-\left\{i^2 \sin ^2 \theta-\cos ^2 \theta\right\}$ $\begin{aligned} & =-\left(-\sin ^2 \theta-\cos ^2 \theta\right) \\ & =-(-1)=1\end{aligned}$

Asked in: AP EAMCET 2021 (23 Aug Shift 2)

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