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 $\begin{vmatrix}\lambda&sin\,\alpha&cos\,\alpha \\ 1&cos\,\alpha &sin\,\alpha \\ -1&sin\,\alpha &-cos\,\alpha \end{vmatrix} = 0$
$\Rightarrow \quad\lambda \left(- cos^{2}\alpha - sin^{2}\alpha \right) - \left(-sin\alpha\, cos\alpha - sin\alpha \, cos\alpha \right) - \left(sin^{2}\alpha - cos^{2}\alpha \right) = 0$
$\Rightarrow \quad -\lambda + sin\, 2\alpha + cos \,2\alpha = 0 \Rightarrow \lambda = sin\, 2\alpha + cos \,2\alpha $
$\Rightarrow \quad\lambda = \sqrt{2}\,cos\left(2\alpha - \frac{\pi}{4}\right)$
Since $-1 < cos \left(2\alpha - \frac{\pi }{4}\right) < 1 \forall \in R$
$\therefore \quad-\sqrt{2} < \lambda \sqrt{2}\, i.e. \,\lambda \in\left[-\sqrt{2}, \,\sqrt{2}\right]$