Question

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

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

Answer 1

The correct option is A $-5$

$det\left(A\right)=\left|\begin{array}{ccc}-2& 4+d& \left(\sin \theta \right)-2\\ 1& \left(\sin \theta \right)+2& d\\ 5& \left(2\sin \theta \right)-d& (-\sin \theta )+2+2d\end{array}\right|$
Apply,
$R_1\to R_1+R_3-2R_2$
$det\left(A\right)=\left|\begin{array}{ccc}1& 0& 0\\ 1& \left(\sin \theta \right)+2& d\\ 5& \left(2\sin \theta \right)-d& (-\sin \theta )+2+2d\end{array}\right|$
$det\left(A\right)=1\left[\right(\sin \theta +2\left)\right(2+2d-\sin \theta )-d(2\sin \theta -d)]-0-0$
$=d^2+4d+4-{\sin }^2\theta$
$\Rightarrow |A|=(d+2{)}^2-{\sin }^2\theta$
$\therefore$ minimum value of $|A|$ occur when ${\sin }^2\theta =1$
$|A{|}_{\text{min}}=(d+2{)}^2-1=8$
$\Rightarrow d^2+4d+3=8$
$\Rightarrow d^2+4d-5=0$
$\Rightarrow (d-1)(d+5)=0$
$\Rightarrow d=1$ or $d=-5$