Question

Consider the system of linear equations in $x,y$and $z$:
$\left(\sin 3\theta \right)x-y+z=0$
$\left(\cos 2\theta \right)x+4y+3z=0$
$2x+7y+7z=0$
The values of $\theta$ for which the system of equations has a non-trivial solution are

• A. $\{n\pi :n\in I\}$
• B. $\{m\pi +(-1{)}^m\pi /6:m\in I\}$
• C. $\{n\pi +(-1{)}^n\pi /6:n\in I\}$
• D. None of these

Answer 1

Answer: (A), (B)

$\{n\pi :n\in I\}$
$\{m\pi +(-1{)}^m\pi /6:m\in I\}$

The given system of equations will have
a non-trivial solution if
$\Delta =\left|\begin{array}{ccc}\sin 3\theta & -1& 1\\ \cos 2\theta & 4& 3\\ 2& 7& 7\end{array}\right|=0$

Applying $R_2\to R_2+4R_1$ and $R_3\to R_3+7R_1$ we get

$\Delta =\left|\begin{array}{ccc}\sin 3\theta & -1& 1\\ \cos 2\theta +4\sin 3\theta & 0& 7\\ 2+7\sin 3\theta & 0& 14\end{array}\right|=0$

Expanding along $C_2$ we get

$\Delta =7[2(\cos 2\theta +4\sin 3\theta )-(2+7\sin 3\theta )]=0$

$\Rightarrow 2\cos 2\theta +\sin 3\theta -2=0$

$2(\cos 2\theta -1)+\sin 3\theta =0$

$-4{\sin }^2\theta +3\sin \theta -4{\sin }^3\theta =0$
$\Rightarrow -\sin \theta (2\sin \theta -1)(2\sin \theta +3)=0$
As $\sin \theta +3$ can never be zero, we get
$\sin \theta =0or\sin \theta =1/2=\sin \left(\pi /6\right)$
$\Rightarrow \theta =n\pi or\theta =m\pi +(-1{)}^m\pi /6(n,m\in I)$