Solve this following

Question:

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

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

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

  1. $-7$

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

  3. $-5$

  4. $2(\sqrt{2}+1)$


Correct Option: , 3

Solution:

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

$\left(\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\right)$

$=\left|\begin{array}{ccc}1 & 0 & 0 \\ 1 & \sin \theta+2 & d \\ 5 & 2 \sin \theta-d & 2+2 d-\sin \theta\end{array}\right|$

$=(2+\sin \theta)(2+2 \mathrm{~d}-\sin \theta)-\mathrm{d}(2 \sin \theta-\mathrm{d})$

$=4+4 \mathrm{~d}-2 \sin \theta+2 \sin \theta+2 \mathrm{~d} \sin \theta-\sin ^{2} \theta-2 \mathrm{~d} \sin \theta+\mathrm{d}^{2}$

$=\mathrm{d}^{2}+4 \mathrm{~d}+4-\sin ^{2} \theta$

$=(\mathrm{d}+2)^{2}-\sin ^{2} \theta$

For a given $d$, minimum value of

$\operatorname{det}(A)=(d+2)^{2}-1=8$

$\Rightarrow d=1$ or $-5$

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