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

Answer 1

Answer: (C)

$-5$

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

$(R_1\to R_1+R_3-2R_2)$

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

$=(2+\sin \theta )(2+2d-\sin \theta )-d(2\sin \theta -d)$

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

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

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

For a given d, minimum value of $det\left(A\right)=(d+2{)}^2-1=8$

$\Rightarrow (d+2{)}^2=9$

$d=3-2=1$ or $d=-3-2=-5$

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