Question

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

• A. $-7$
• B. $2 ( \sqrt{2} + 2)$
• C. $-5$
• D. $2 ( \sqrt{2} + 1)$

Answer 1

Answer: (C)

$$
\operatorname{det} A=\left|\begin{array}{ccc}
-2 & 4+d & \sin \theta-2 \\
1 & \sin \theta+2 & d \\
5 & 2 \sin \theta-d & -\sin \theta+2+2 d
\end{array}\right|
$$
$\left( R _{1} \rightarrow R _{1}+ R _{3}-2 R _{2}\right)$
$$
=\left|\begin{array}{ccc}
1 & 0 & 0 \\
1 & \sin \theta+2 & d \\
5 & 2 \sin \theta-d & 2+2 d-\sin \theta
\end{array}\right|
$$
$=(2+\sin \theta)(2+2 d -\sin \theta)- d (2 \sin \theta- d )$
$=4+4 d -2 \sin \theta+2 \sin \theta+2 dsin \theta-\sin ^{2} \theta-2 d \sin \theta+ d ^{2}$
$= d ^{2}+4 d +4-\sin ^{2} \theta$
$=(d+2)^{2}-\sin ^{2} \theta$
For a given d, minimum value of
$\operatorname{det}( A )=( d +2)^{2}-1=8$
$\Rightarrow d =1$ or $-5$