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(D) 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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$(1.5, 3)$

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(D) 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 + \sin^2 	heta - 0 + \sin^2 	heta + 1 = 2 + 2\sin^2 	heta = 2(1 + \sin^2 	heta)$Given $	heta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} ight)$,the value of $\sin 	heta$ ranges from $-\frac{1}{\sqrt{2}}$ to $\frac{1}{\sqrt{2}}$.Specifically,$-\frac{1}{\sqrt{2}} Squaring the inequality,we get $0 \le \sin^2 	heta Now,substitute this into the expression for $|A|$:$|A| = 2(1 + \sin^2 	heta)$Since $0 \le \sin^2 	heta Multiplying by $2$,we get $2 \le 2(1 + \sin^2 	heta) Thus,$\det(A) \in [2, 3)$.Comparing this with the given options,the interval $[2, 3)$ is contained within $(1.5, 3)$.

If $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,then for all $\theta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$,$\det(A)$ lies in the interval:

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