Let mathrm{A}=left[begin{array}{ccc}1 & s | vedclass

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Question

$A=\left[\begin{array}{ccc}1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1\end{array}ight]$

Vedclass · Accepted Answer

$\operatorname{Det}(\mathrm{A}) \in[2,4]$

Vedclass · Answer

$A=\left[\begin{array}{ccc}1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1\end{array}ight]$

$	herefore|A|=1\left(1+\sin ^{2} 	hetaight)-\sin 	heta(-\sin 	heta+\sin 	heta)+1\left(\sin ^{2} 	heta+1ight)$

$=1+\sin ^{2} 	heta+\sin ^{2} 	heta+1$

$=2+2 \sin ^{2} 	heta$

$=2\left(1+\sin ^{2} 	hetaight)$

Now, $0 \leq 	heta \leq 2 \pi$

$\Rightarrow 0 \leq \sin 	heta \leq 1$

$\Rightarrow 0 \leq \sin ^{2} 	heta \leq 1$

$\Rightarrow 1 \leq 1+\sin ^{2} 	heta \leq 2$

$\Rightarrow 2 \leq 2\left(1+\sin ^{2} 	hetaight) \leq 4$

$	herefore \operatorname{Det}(A) \in[2,4]$

Hence, the correct answer is $B$.

Let $\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right],$ where $0 \leq \theta \leq 2 \pi$. Then

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