Matrices and Determinants 2 Question 9
11. Let $d \in R$, and
$A=$ $\begin{vmatrix} -2 & 4+d & (\sin \theta)-2\\ 1 & (\sin \theta)+2 & d\\ 5 & (2 \sin \theta)-d & (-\sin \theta)+2+2 d \end{vmatrix}$ ,$\theta \in[\theta, 2 \pi]$.
If the minimum value of $\operatorname{det}(\mathrm{A})$ is 8 , then a value of $d$ is
(2019 Main, 10 Jan I)
(a) -5
(b) -7
(c) $2(\sqrt{2}+1)$
(d) $2(\sqrt{2}+2)$
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Answer:
Correct Answer: 11. (a)
Solution:
- Given,
$ A=\begin{vmatrix} -2 & 4+d & (\sin \theta)-2 \\ 1 & (\sin \theta)+2 & d \\ 5 & (2 \sin \theta)-d & (-\sin \theta)+2+2 d \end{vmatrix} $
$ \begin{aligned} \therefore \quad|A| & =\begin{vmatrix} -2 & 4+d & (\sin \theta)-2 \\ 1 & (\sin \theta)+2 & d \\ 5 & (2 \sin \theta)-d & (-\sin \theta)+2+2 d \end{vmatrix} \\ & =\begin{vmatrix} -2 & 4+d & (\sin \theta)-2 \\ 1 & (\sin \theta)+2 & d \\ 1 & 0 & 0 \end{vmatrix} \\ & =1[(4+d) d-(\sin \theta+2)(\sin \theta-2)] \\ & =\left(d^{2}+4 d-\sin ^{2} \theta+4\right) \\ & =\left(d^{2}+4 d+4\right)-\sin ^{2} \theta \\ & =(d+2)^{2}-\sin ^{2} \theta \end{aligned} $
Note that $|A|$ will be minimum if $\sin ^{2} \theta$ is maximum i.e. if $\sin ^{2} \theta$ takes value 1 .
$ \begin{aligned} & \because \quad|A|_{\min }=8, \\ & \text { therefore }(d+2)^{2}-1=8 \\ & \Rightarrow \quad(d+2)^{2}=9 \\ & \Rightarrow \quad d+2= \pm 3 \\ & \Rightarrow \quad d=1,-5 \end{aligned} $
