Question

The value of $\theta $ for which the system of equations

$\left(\left(sin 3 \theta \right) x - 2 y + 3 z = 0, \left(\right. cos ⁡ 2 \theta \, \right)x+8y-7z=0$ and $2x+14y-11z=0$ has a non-trivial solution, is (here, $n\in Z$ )

• A. $n\pi $
• B. $n\pi \, + \, \left(- 1\right)^{n} \, \pi / 3$
• C. $n\pi \, + \, \left(- 1\right)^{n} \, \pi / 8$
• D. None of these

Answer 1

Answer: (A)

The system of equations has a non-trivial solution if and only if $\begin{vmatrix} \, sin 3 \theta & -2 & 3 \\ cos ⁡ 2 \theta & 8 & -7 \\ 2 & 14 & -11 \, \end{vmatrix}= \, 0$
Applying $R_{2} \rightarrow \, R_{2} \, + \, 4R_{1}, \, R_{3} \rightarrow \, R_{3} \, + \, 7R_{1}$ , we get,
$\begin{vmatrix} \, sin 3 \theta & -2 & 3 \\ cos ⁡ 2 \theta +4sin ⁡ 3 \theta & 0 & 5 \\ 2+7sin ⁡ 3 \theta & 0 & 10 \, \end{vmatrix}= \, 0$
Expanding along $C_{2}$ , we get,
$2\left(cos 2 \theta \, + \, 4 sin ⁡ 3 \theta \right)- \, \left(2 \, + \, 7 sin ⁡ 3 \theta \right)=0$
$\Rightarrow \, 2- \, 2 \, cos 2 \theta - \, sin ⁡ 3 \theta =0$
$\Rightarrow 4\left(sin\right)^{2} \theta - \, \left(3 sin ⁡ \theta - \, 4 \left(sin\right)^{3} ⁡ \theta \right)=0$
$\Rightarrow sin \theta \, \left(4 \left(sin\right)^{2} ⁡ \theta \, + \, 4 sin ⁡ \theta - \, 3\right)=0$
$\Rightarrow sin \theta \, \left(2 sin ⁡ \theta - \, 1\right) \, \left(2 sin ⁡ \theta \, + \, 3\right)=0$
$\Rightarrow sin \theta =0$ or $sin \theta =1 / 2$ .
[ $sin \theta =- 3 / 2$ is not possible]
$\therefore $ For, $\theta =n\pi $ the system of equations has a non-trivial solution.