Question

Let S denote the set of all values of $\lambda$ for which the system of linear equations
$\lambda x+\left(\sin \alpha \right)y+\left(\cos \alpha \right)z=0$,
$x+\left(\cos \alpha \right)y+\left(\sin \alpha \right)z=0$,
$-x+\left(\sin \alpha \right)y-\left(\cos \alpha \right)z=0$,
has a non-trivial solution $\forall \alpha \in R$, then S contains

• A. $(-1,1)$
• B. $[-\sqrt{2},-1]$
• C. $[1,\sqrt{2}]$
• D. $(-2,2)$

Answer 1

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

The correct options are
A $(-1,1)$
B $[-\sqrt{2},-1]$
C $[1,\sqrt{2}]$

For non-trivial solution,
$\left|\begin{array}{ccc}\lambda & \sin \alpha & \cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ -1& \sin \alpha & -\cos \alpha \end{array}\right|=0$
Expanding along $C_1$
$\Rightarrow \lambda (-{\cos }^2\alpha -{\sin }^2\alpha )-1(-\sin \alpha \cos \alpha -\sin \alpha \cos \alpha )-1({\sin }^2\alpha -{\cos }^2\alpha )=0$
$\Rightarrow -\lambda +\sin 2\alpha +\cos 2\alpha =0$
$\Rightarrow \lambda =\sin 2\alpha +\cos 2\alpha =\sqrt{2}\sin (\frac{\pi }{4}+2\alpha )$
Since, $-1\le \sin (\frac{\pi }{4}+2\alpha )\le 1$
$\Rightarrow -\sqrt{2}\le \lambda \le \sqrt{2}$