Matrices and Determinants 2 Question 1
1. A value of $\theta \in(0, \pi / 3)$, for which
$\left|\begin{matrix}1+\cos ^{2} \theta & \sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & 1+\sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & \sin ^{2} \theta & 1+4 \cos 6 \theta\end{matrix}\right|$
(a) $\frac{\pi}{9}$
(b) $\frac{\pi}{18}$
(c) $\frac{7 \pi}{24}$
(d) $\frac{7 \pi}{36}$
2019 Main, 12 April II
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Answer:
Correct Answer: 1. (a)
Solution:
- Let $\Delta=\left|\begin{matrix}1+\cos ^{2} \theta & \sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & 1+\sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & \sin ^{2} \theta & 1+4 \cos 6 \theta\end{matrix}\right|=0$
Applying $C_{1} \rightarrow C_{1}+C_{2}$, we get
$ \Delta=\left|\begin{matrix} 2 & \sin ^{2} \theta & 4 \cos 6 \theta \\ 2 & 1+\sin ^{2} \theta & 4 \cos 6 \theta \\ 1 & \sin ^{2} \theta & 1+4 \cos 6 \theta \end{matrix}\right|=0 $
Applying $R_{1} \rightarrow R_{1}-2 R_{3}$ and $R_{2} \rightarrow R_{2}-2 R_{3}$, we get
$ \Delta=\left|\begin{matrix} 0 & -\sin ^{2} \theta & -2-4 \cos 6 \theta \\ 0 & 1-\sin ^{2} \theta & -2-4 \cos 6 \theta \\ 1 & \sin ^{2} \theta & 1+4 \cos 6 \theta \end{matrix}\right|=0 $
On expanding w.r.t. $C_{1}$, we get
$\Rightarrow \sin ^{2} \theta(2+4 \cos 6 \theta)+(2+4 \cos 6 \theta)\left(1-\sin ^{2} \theta\right)=0$
$\Rightarrow 2+4 \cos 6 \theta=0 \Rightarrow \cos 6 \theta=-\frac{1}{2}=\cos \frac{2 \pi}{3}$
$ \Rightarrow \quad 6 \theta=\frac{2 \pi}{3} \Rightarrow \theta=\frac{\pi}{9} \quad $
$\because \theta \in 0, \frac{\pi}{3} $
