If A(theta)=begin{bmatrix} i sin theta & cos | vedclass

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(C) Given $A(	heta)=\begin{bmatrix} i \sin 	heta & \cos 	heta \ \cos 	heta & i \sin 	heta \end{bmatrix}$.The determinant is $\operatorname{det}

Vedclass · Accepted Answer

$\operatorname{det}[A(	heta)]^{-1}=1$

Vedclass · Answer

(C) Given $A(	heta)=\begin{bmatrix} i \sin 	heta & \cos 	heta \ \cos 	heta & i \sin 	heta \end{bmatrix}$.The determinant is $\operatorname{det} A(	heta) = (i \sin 	heta)(i \sin 	heta) - (\cos 	heta)(\cos 	heta) = i^2 \sin^2 	heta - \cos^2 	heta$.Since $i^2 = -1$,we have $\operatorname{det} A(	heta) = -\sin^2 	heta - \cos^2 	heta = -(\sin^2 	heta + \cos^2 	heta) = -1$.Since the determinant is a constant $-1$ regardless of $	heta$,we have $\operatorname{det} A(\pi+	heta) = -1$,$\operatorname{det} A(-	heta) = -1$,and $\operatorname{det} A(	heta) = -1$.Checking the options:Option $A$: $\operatorname{det} A(\pi+	heta) = -1$ and $\operatorname{det} A(-	heta) = -1$. So,$-1 = -1$ (True).Option $B$: $\operatorname{det} A(-	heta) = -1$ and $\operatorname{det} A(	heta) = -1$. So,$-1 = -1$ (True).Option $C$: $\operatorname{det}[A(	heta)]^{-1} = \frac{1}{\operatorname{det} A(	heta)} = \frac{1}{-1} = -1$. The statement says $1$,which is False.Option $D$: $\operatorname{det} A(-	heta) = -1$ (True).Therefore,the statement that is not true is $C$.

If $A(\theta)=\begin{bmatrix} i \sin \theta & \cos \theta \\ \cos \theta & i \sin \theta \end{bmatrix}$ is a matrix,where $i=\sqrt{-1}$,then which of the following is not true?

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