Question

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

$\left(\left(sin3\theta \right)x-2y+3z=0,(cos2\theta \right)x+8y-7z=0$ 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 $\left|\begin{array}{ccc}sin3\theta & -2& 3\\ cos2\theta & 8& -7\\ 2& 14& -11\end{array}\right|=0$
Applying $R_2\to R_2+4R_1,R_3\to R_3+7R_1$ , we get,
$\left|\begin{array}{ccc}sin3\theta & -2& 3\\ cos2\theta +4sin3\theta & 0& 5\\ 2+7sin3\theta & 0& 10\end{array}\right|=0$
Expanding along $C_2$ , we get,
$2\left(cos2\theta +4sin3\theta \right)-\left(2+7sin3\theta \right)=0$
$\Rightarrow 2-2cos2\theta -sin3\theta =0$
$\Rightarrow 4{\left(sin\right)}^2\theta -\left(3sin\theta -4{\left(sin\right)}^3\theta \right)=0$
$\Rightarrow sin\theta \left(4{\left(sin\right)}^2\theta +4sin\theta -3\right)=0$
$\Rightarrow sin\theta \left(2sin\theta -1\right)\left(2sin\theta +3\right)=0$
$\Rightarrow sin\theta =0$ 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.