Let and,

Question:

Let $d \in \mathbf{R}$, and

$A=\left[\begin{array}{ccc}-2 & 4+d & (\sin \theta)-2 \\ 1 & (\sin \theta)+2 & d \\ 5 & (2 \sin \theta)-d & (-\sin \theta)+2+2 d\end{array}\right]$

$\theta \in[0,2 \pi] .$ If the minimum value of $\operatorname{det}(\mathrm{A})$ is 8, then a value of $d$ is:

1. (1) $-5$

2. (2) $-7$

3. (3) $2(\sqrt{2}+1)$

4. (4) $2(\sqrt{2}+2)$

Correct Option: 1

Solution:

$\operatorname{det}(A)=\left|\begin{array}{ccc}-2 & 4+d & \sin \theta-2 \\ 1 & \sin \theta+2 & d \\ 5 & 2 \sin \theta-d & -\sin \theta+2+2 d\end{array}\right|$

Applying $R_{3} \rightarrow R_{3}-2 R_{2}+R_{1}$ we get

$\operatorname{det}(A)=\left|\begin{array}{ccc}-2 & 4+d & \sin \theta-2 \\ 1 & \sin \theta+2 & d \\ 1 & 0 & 0\end{array}\right|$

$=d(4+d)-\left(\sin ^{2} \theta-4\right)$

$\Rightarrow \operatorname{det}(A)=d^{2}+4 d+4-\sin ^{2} \theta=(d+2)^{2}-\sin ^{2} \theta$

Minimum value of $\operatorname{det}(A)$ is attained when $\sin ^{2} \theta=1$

$\therefore(d+2)^{2}-1=8 \Rightarrow(d+2)^{2}=9 \Rightarrow d+2$

$=\pm 3$

$\Rightarrow d=-5$ or 1