Question

The system of equations
$x-y\cos \theta +z\cos 2\theta =0$
$-x\cos \theta +y-z\cos \theta =0$
$x\cos 2\theta -y\cos \theta +z=0$
has non trivial solution for $\theta$ equals to

• A. $\pi /3$
• B. $\pi /6$
• C. $2\pi /3$
• D. $\pi /12$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\pi /3$
B $\pi /6$
C $2\pi /3$
D $\pi /12$

The coefficient matrix for the given system of linear equations is given by-

$A=\left[\begin{array}{ccc}1& -\cos \theta & \cos 2\theta \\ -\cos \theta & 1& -\cos \theta \\ \cos 2\theta & -\cos \theta & 1\end{array}\right]$

This system of equations will have a non trivial solution if $detA=0$

Now, $detA=\left|\begin{array}{ccc}1& -\cos \theta & \cos 2\theta \\ -\cos \theta & 1& -\cos \theta \\ \cos 2\theta & -\cos \theta & 1\end{array}\right|$

$\Rightarrow detA=\left|\begin{array}{ccc}1-\cos 2\theta & -\cos \theta & \cos 2\theta \\ 0& 1& -\cos \theta \\ \cos 2\theta -1& -\cos \theta & 1\end{array}\right|$ ($C_1\to C_1-C_3$)

$\Rightarrow detA=\left|\begin{array}{ccc}1-\cos 2\theta & -\cos \theta & \cos 2\theta \\ 0& 1& -\cos \theta \\ 0& -2\cos \theta & 1+\cos \theta \end{array}\right|$ ($R_3\to R_3+R_1$)

Expanding along $C_1$ we get-

$detA=(1-\cos 2\theta )\left\{\right(1+\cos \theta )-(2{\cos }^2\theta \left)\right\}$

$\Rightarrow detA=(1-\cos 2\theta )(2{\cos }^2\theta -2{\cos }^2\theta )$ (Since, $1+\cos 2\theta =2{\cos }^2\theta$)

$\Rightarrow detA=0$

This implies that $detA=0$ irrespective of the value of $\theta$. Hence the value of the determinant is 0 for all values of $\theta$.

Thus, the given system of linear equations has a non trivial solution for every value of $\theta$.

Hence, the correct answer is given by all four options, i.e.- options A,B,C,D are all correct.