Question

Let $\lambda$ and $\alpha$ be real. 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 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 $[1,\sqrt{2}]$

The given system of linear equations will have a non-trivial solution if
$\left|\begin{array}{ccc}\lambda & \sin \alpha & \cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ -1& \sin \alpha & -\cos \alpha \end{array}\right|=0$
Expanding the determinant along $C_1$ we get
$\lambda (-{\cos }^2\alpha -{\sin }^2\alpha )-(-\sin \alpha \cos \alpha -\sin \alpha \cos \alpha )-({\sin }^2\alpha -{\cos }^2\alpha )=0$
$\Rightarrow -\lambda +\sin 2\alpha +\cos 2\alpha =0$
$\Rightarrow \lambda =\sin 2\alpha +\cos 2\alpha =\sqrt{2}\sin (\pi /4+2\alpha )$
$\Rightarrow -\sqrt{2}\le \lambda \le \sqrt{2}$
Hence, possible options are A,B and C.