Let d in mathbb{R},and A = begin{bmatrix} -2 | vedclass

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(A) Given the determinant $A = \begin{vmatrix} -2 & 4+d & \sin 	heta - 2 \ 1 & \sin 	heta + 2 & d \ 5 & 2\sin 	heta - d & -\sin 	heta + 2 + 2d \

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$-5$

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(A) Given the determinant $A = \begin{vmatrix} -2 & 4+d & \sin 	heta - 2 \ 1 & \sin 	heta + 2 & d \ 5 & 2\sin 	heta - d & -\sin 	heta + 2 + 2d \end{vmatrix}$.Applying the row operation $R_3 	o R_3 + 2R_2 + R_1$:$R_3 = [5 + 2(1) + (-2), (2\sin 	heta - d) + 2(\sin 	heta + 2) + (4 + d), (-\sin 	heta + 2 + 2d) + 2(d) + (\sin 	heta - 2)]$$R_3 = [5, 4\sin 	heta + 8, 4d] = 5[1, \sin 	heta + 2, d]$.Since $R_3$ is a multiple of $R_2$,the determinant is $0$ if we perform $R_3 	o R_3 - 5R_2$. Wait,let us re-evaluate the determinant directly.Expanding along $R_1$: $\det(A) = -2[(\sin 	heta + 2)(-\sin 	heta + 2 + 2d) - d(2\sin 	heta - d)] - (4+d)[1(-\sin 	heta + 2 + 2d) - 5d] + (\sin 	heta - 2)[1(2\sin 	heta - d) - 5(\sin 	heta + 2)]$.After simplification,$\det(A) = (d+2)^2 - \sin^2 	heta$.To find the minimum value of $\det(A)$,we know $\sin^2 	heta \in [0, 1]$.Thus,$\min(\det(A)) = (d+2)^2 - 1$.Given $\min(\det(A)) = 8$,we have $(d+2)^2 - 1 = 8 \implies (d+2)^2 = 9$.$d+2 = 3$ or $d+2 = -3$.$d = 1$ or $d = -5$.

Let $d \in \mathbb{R}$,and $A = \begin{bmatrix} -2 & 4+d & \sin \theta - 2 \\ 1 & \sin \theta + 2 & d \\ 5 & 2\sin \theta - d & -\sin \theta + 2 + 2d \end{bmatrix}$,where $\theta \in [0, 2\pi]$. If the minimum value of $\det(A)$ is $8$,then a value of $d$ is

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