Matrices and Determinants 4 Question 26
27. Let $\lambda$ and $\alpha$ be real. Find the set of all values of $\lambda$ for which the system of linear equations
$ \begin{aligned} \lambda x+(\sin \alpha) y+(\cos \alpha) z & =0 \\ x+(\cos \alpha) y+(\sin \alpha) z & =0 \end{aligned} $
$ \text { and } \quad-x+(\sin \alpha) y-(\cos \alpha) z=0 $
has a non-trivial solution.
For $\lambda=1$, find all values of $\alpha$.
$(1993,5$ M)
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Answer:
Correct Answer: 27. $\theta=n \pi, n \pi+(-1)^{n} \frac{\pi}{6}, n \in Z$
Solution:
- Given, $\lambda x+(\sin \alpha) y+(\cos \alpha) z=0$
$ x+(\cos \alpha) y+(\sin \alpha) z=0 $
and $-x+(\sin \alpha) y-(\cos \alpha) z=0$ has non-trivial solution.
$ \begin{aligned} & \therefore \quad \Delta=0 \\ & \begin{array}{l} \Rightarrow \quad\left|\begin{array}{ccc} \lambda & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ -1 & \sin \alpha & -\cos \alpha \end{array}\right|=0 \\ \Rightarrow \quad \lambda\left(-\cos ^{2} \alpha-\sin ^{2} \alpha\right)-\sin \alpha(-\cos \alpha+\sin \alpha) \end{array} \\ & \Rightarrow \quad-\lambda+\sin \alpha \cos \alpha+\sin \alpha \cos \alpha-\sin ^{2} \alpha+\cos ^{2} \alpha=0 \\ & \Rightarrow \quad \lambda=\cos 2 \alpha+\sin 2 \alpha \\ & \because-\sqrt{a^{2}+b^{2}} \leq a \sin \theta+b \cos \theta \leq \sqrt{a^{2}+b^{2}} \\ & \therefore \quad-\sqrt{2} \leq \lambda \leq \sqrt{2} \end{aligned} $
Again, when $\lambda=1, \cos 2 \alpha+\sin 2 \alpha=1$
$ \begin{aligned} & \Rightarrow \quad \frac{1}{\sqrt{2}} \cos 2 \alpha+\frac{1}{\sqrt{2}} \sin 2 \alpha=\frac{1}{\sqrt{2}} \\ & \Rightarrow \quad \cos (2 \alpha-\pi / 4)=\cos \pi / 4 \\ & \therefore \quad 2 \alpha-\pi / 4=2 n \pi \pm \pi / 4 \\ & \Rightarrow \quad 2 \alpha=2 n \pi-\pi / 4+\pi / 4 \text { or } 2 \alpha=2 n \pi+\pi / 4+\pi / 4 \\ & \therefore \quad \alpha=n \pi \text { or } n \pi+\pi / 4 \end{aligned} $
