Let A = begin{bmatrix} 1 & sin theta & 1 -si | vedclass

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(B) Given $A = \begin{bmatrix} 1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1 \end{bmatrix}$.Expanding the determinant

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

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(B) Given $A = \begin{bmatrix} 1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1 \end{bmatrix}$.Expanding the determinant along the first row:$|A| = 1(1 - (-\sin^2 	heta)) - \sin 	heta(-\sin 	heta - (-\sin 	heta)) + 1(\sin^2 	heta - (-1))$$|A| = 1(1 + \sin^2 	heta) - \sin 	heta(0) + 1(\sin^2 	heta + 1)$$|A| = 1 + \sin^2 	heta + \sin^2 	heta + 1$$|A| = 2 + 2 \sin^2 	heta = 2(1 + \sin^2 	heta)$.Given $0 \leq 	heta \leq 2 \pi$,the range of $\sin 	heta$ is $[-1, 1]$.Therefore,the range of $\sin^2 	heta$ is $[0, 1]$.Adding $1$ to the inequality: $1 \leq 1 + \sin^2 	heta \leq 2$.Multiplying by $2$: $2 \leq 2(1 + \sin^2 	heta) \leq 4$.Thus,$\operatorname{Det}(A) \in [2, 4]$.Hence,the correct answer is $B$.

Let $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,where $0 \leq \theta \leq 2 \pi$. Then

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