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

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(C) The determinant of matrix $A$ is given by:$|A| = \begin{vmatrix} 1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1 \e

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

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(C) The determinant of matrix $A$ is given by:$|A| = \begin{vmatrix} 1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1 \end{vmatrix}$Expanding 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 = 2(1 + \sin^2 	heta)$Since we know that $-1 \le \sin 	heta \le 1$,it follows that $0 \le \sin^2 	heta \le 1$.Adding $1$ to all parts,we get $1 \le 1 + \sin^2 	heta \le 2$.Multiplying by $2$,we get $2 \le 2(1 + \sin^2 	heta) \le 4$.Therefore,$|A| \in [2, 4]$.

Let $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,where $0 \le \theta < 2\pi$,then:

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