Question

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

• A. $\alpha =n\pi$
• B. $\alpha =n\frac{\pi }{2}$
• C. $n\pi +\frac{\pi }{4}$
• D. $n\frac{\pi }{2}+\frac{\pi }{4}$

Answer 1

Answer: (A), (C)

$n\pi +\frac{\pi }{4}$

Given, $\lambda x+\left(sin\alpha \right)y+\left(cos\alpha \right)z=0,x+\left(cos\alpha \right)y+\left(sin\alpha \right)z=0and-x+\left(sin\alpha \right)y-\left(cos\alpha \right)z=0$ has non - trivial solution.
$\therefore \Delta =0$
$\Rightarrow \left|\begin{array}{ccc}\lambda & sin\alpha & cos\alpha \\ 1& cos\alpha & sin\alpha \\ -1& sin\alpha & -cos\alpha \end{array}\right|=0$
$\Rightarrow \lambda (-co{s}^2\alpha -si{n}^2\alpha )-sin\alpha (-cos\alpha +sin\alpha )+cos\alpha (sin\alpha +cos\alpha )=0\Rightarrow -\lambda +sin\alpha cos\alpha +sin\alpha cos\alpha -si{n}^2\alpha +co{s}^2\alpha =0\Rightarrow \lambda =cos2\alpha +sin2\alpha [\because -\sqrt{a^2+b^2}\le asin\theta +bcos\theta \le \sqrt{a^2+b^2}]$
$\therefore -\sqrt{2}\le \lambda \le \sqrt{2}$ . . . (i)
Again, when $\lambda =1$, $cos2\alpha +sin2\alpha =1$
$\Rightarrow \frac{1}{\sqrt{2}}cos2\alpha +\frac{1}{\sqrt{2}}sin2\alpha =\frac{1}{\sqrt{2}}\Rightarrow cos(2\alpha -\frac{\pi }{4})=cos\frac{\pi }{4}\therefore 2\alpha -\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}$
$\Rightarrow 2\alpha =2n\pi -\frac{\pi }{4}+\frac{\pi }{4}or2\alpha =2n\pi +\frac{\pi }{4}+\frac{\pi }{4}\therefore \alpha =n\pi orn\pi +\frac{\pi }{4}$