Let $d \in R $ and

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

-5

The correct answer is option (1) : -5

$A= \begin{bmatrix} -2 & 4 + d& sin \theta - 2\\1 & sin \theta + 2 & d\\5 & 2 sin \theta - d& -sin \theta + 2 + 2d \end {bmatrix}$

$⇒|A|=\begin{vmatrix}-2 & 4 + d & sin \theta - 2\\1 & sin \theta + 2 & d\\1 & 0 & 0 \end{vmatrix}$

Applying $R_3→R_3-2R_2+R_1$

$⇒|A|=(4 + d) d- (sin^2 \theta - 4) = d^2 + 4d + 4 -sin^2 \theta $

$⇒|A|=(d+2)^2 -sin^2\theta $ ...............(i)

Clearly, |A| is minimum when sin $\theta = \pm 1.$

Substituting $|A|=8 $ and $sin \theta = \pm 1 $ in (i) ,we get

$∴8= (d+2)^2 -1⇒d= -5, 1$