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