Let $\lambda$ and $\alpha$ be real. If the range of the set of all values of $\lambda$ for which the

Question

Let $\lambda$ and $\alpha$ be real. If the range of the set of all values of $\lambda$ for which the system of linear equations $\lambda x+(\sin \alpha) y+(\cos \alpha) z=0, x+(\cos \alpha) y+(\sin \alpha) z=0$, $-x+(\sin \alpha) y-(\cos \alpha) z=0$ has a non trivial solution is $ [\alpha, \beta]$ then $\alpha^2 + \beta^2$.

Professor PC Thomas Classes · Answer

$\Delta=\left|\begin{array}{ccc}\lambda & \sin \alpha & \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ -1 & \sin \alpha & -\cos \alpha\end{array}ight|=0$ Expanding along first column, we get\ $\lambda\left[-\left(\cos ^{2} \alpha+\sin ^{2} \alphaight)ight]-1(-\sin \alpha \cos \alpha-\sin \alpha$ $\cos \alpha)-1\left(\sin ^{2} \alpha-\cos ^{2} \alphaight)=0$\ $\Rightarrow \lambda(-1)-1(-2 \sin \alpha \cos \alpha)$\ $-\left(\sin ^{2} \alpha-\cos ^{2} \alphaight)=0$\ $\Rightarrow \quad \lambda=\cos ^{2} \alpha-\sin ^{2} \alpha+2 \sin \alpha \cos \alpha$\ $\Rightarrow \quad \lambda=\cos 2 \alpha+\sin 2 \alpha$\ $\left(\because \cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha, 2 \sin \alphaight.$\ $\cos \alpha=\sin 2 \alpha)$\ If $\lambda=1$, then $\cos 2 \alpha+\sin 2 \alpha=1$\ $\Rightarrow \frac{1}{\sqrt{2}} \cos 2 \alpha+\frac{1}{\sqrt{2}} \sin 2 \alpha=\frac{1}{\sqrt{2}}$\ $\Rightarrow \quad \cos \left(2 \alpha-\frac{\pi}{4}ight)=\cos \frac{\pi}{4}$\ $	herefore \quad 2 \alpha-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4}$\ Hence, $2 \alpha=2 n \pi+\frac{\pi}{4}+\frac{\pi}{4}, n \in Z$\ $\alpha=n \pi$ or $n \pi+\frac{\pi}{4}, n \in Z$.\