Question

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

• A. $\frac{m\pi }{2}$
• B. $m\pi$
• C. $n\pi +{(-1)}^n\frac{\pi }{6}$
• D. None of these

Answer 1

Answer: (B), (C)

The correct options are
B $m\pi$
C $n\pi +{(-1)}^n\frac{\pi }{6}$

Given a system of equations can be written as $AX=0$

where $A=\left[\begin{array}{ccc}\sin 3\theta & -1& 1\\ \cos 2\theta & 4& 3\\ 2& 7& 7\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],Y=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
For non trivial solution,

$|A|=0$

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

$C_2\to C_2+C_3$

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

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

$\Rightarrow \sin \theta (4{\sin }^2\theta +4\sin \theta -3)=0$

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

$\Rightarrow \sin \theta =0or\sin \theta =\frac{1}{2}(\because \sin \theta =-\frac{3}{2}$ (not possible))

$\theta =m\pi$ or $\theta =n\pi +{(-1)}^n\frac{\pi }{6}$