Question

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

• A. $\left[0,\sqrt{2}\right]$
• B. $\left[-\sqrt{2},0\right]$
• C. $\left[-\sqrt{2},\sqrt{2}\right]$
• D. None of these

Answer 1

Answer: (C)

Since the system has a non-trivial solution,
therefore $\left|\begin{array}{ccc}\lambda & sin\alpha & cos\alpha \\ 1& cos\alpha & sin\alpha \\ -1& sin\alpha & -cos\alpha \end{array}\right|=0$
$\Rightarrow \lambda \left(-co{s}^2\alpha -si{n}^2\alpha \right)-\left(-sin\alpha cos\alpha -sin\alpha cos\alpha \right)-\left(si{n}^2\alpha -co{s}^2\alpha \right)=0$
$\Rightarrow -\lambda +sin2\alpha +cos2\alpha =0\Rightarrow \lambda =sin2\alpha +cos2\alpha$
$\Rightarrow \lambda =\sqrt{2}cos\left(2\alpha -\frac{\pi }{4}\right)$
Since $-1<cos\left(2\alpha -\frac{\pi }{4}\right)<1\forall \in R$
$\therefore -\sqrt{2}<\lambda \sqrt{2}i.e.\lambda \in \left[-\sqrt{2},\sqrt{2}\right]$