Question

Let $\lambda$ and $\alpha$ be real. Find the set of all values of $\lambda$ for which the system of linear equation
$\lambda x+\left(sin\alpha \right)y+\left(cos\alpha \right)z=0$
$x+\left(cos\alpha \right)y+\left(sin\alpha \right)z=0$
$-x+\left(sin\alpha \right)y-\left(cos\alpha \right)z=0$ has a non-trivial solution. For $\lambda =1$, find all values of $\alpha$

• A. $\lambda =sin2\alpha +cos2\alpha ;\alpha =m\pi$ or $\alpha =k\pi +\pi /4(m,k\epsilon I)$
• B. $\lambda =sin\alpha +cos\alpha ;\alpha =m\pi$ or $\alpha =k\pi +\pi /4(m,k\epsilon I)$
• C. $\lambda =sin\alpha +cos2\alpha ;\alpha =m\pi$ or $\alpha =k\pi +\pi /4(m,k\epsilon I)$
• D. $\lambda =sin2\alpha +cos\alpha ;\alpha =m\pi$ or $\alpha =k\pi +\pi /4(m,k\epsilon I)$

Answer 1

The correct option is A $\lambda =sin2\alpha +cos2\alpha ;\alpha =m\pi$ or $\alpha =k\pi +\pi /4(m,k\epsilon I)$

Given $\lambda x+\left(\sin \alpha \right)y+\left(\cos \alpha \right)z=0,x+\left(\cos \alpha \right)y+\left(\sin \alpha \right)z=0$ and $-x+\left(\sin \alpha \right)y-\left(\cos \alpha \right)z=0$
has non trivial solution
$\therefore △=0\Rightarrow \left|\begin{array}{ccc}\lambda & \sin \alpha & \cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ -1& \sin \alpha & -\cos \alpha \end{array}\right|=0$
$\Rightarrow \lambda (-{\cos }^2\alpha -{\sin }^2\alpha )-\sin \alpha (-\cos \alpha +\sin \alpha )+\cos \alpha (\sin \alpha +\cos \alpha )\Rightarrow -\lambda +\sin \alpha \cos \alpha +\sin \alpha \cos \alpha -{\sin }^2\alpha +{\cos }^2\alpha =0\Rightarrow \lambda =\cos 2\alpha +\sin 2\alpha (\because -\sqrt{a^2+b^2}\le a\sin \theta +b\cos \theta \le \sqrt{a^2+b^2})$
$\therefore -\sqrt{2}\le \lambda \le \sqrt{2}$ ...(1)
When $\lambda =1$
$\cos 2\alpha +\sin 2\alpha =1\Rightarrow \frac{1}{\sqrt{2}}\cos 2\alpha +\frac{1}{\sqrt{2}}\sin 2\alpha =\frac{1}{\sqrt{2}}$
$\Rightarrow \cos (2\alpha -\frac{\pi }{4})=\cos \left(\frac{\pi }{4}\right)$
$\therefore 2\alpha -\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}$
$\Rightarrow 2\alpha =2n\pi -\frac{\pi }{4}+\frac{\pi }{4}$ and $2\alpha =2n\pi +\frac{\pi }{4}+\frac{\pi }{4}$
$\therefore \alpha =n\pi$ and $\alpha =n\pi +\frac{\pi }{4}$