Q103 / 200 · Manipal (MET) 2012 · Mathematics

Let $\lambda$ and $\alpha$ be real $\begin{array}{r} \lambda x+\sin \alpha \cdot y+\cos \alpha \cdot z=0 \\ x+\cos \alpha \cdot y+\sin \alpha \cdot z=0 \\ -x+\sin \alpha \cdot y-\cos \alpha \cdot z=0 \end{array}$ has a non-trivial solution. If $\lambda=1$, then all the values of $\alpha$ is equals

  • A
    $n \pi+\frac{\pi}{2}$
  • Correct answer
    B
    $n \pi+\frac{\pi}{4}, n \pi$
  • C
    $n \pi \pm \frac{\pi}{2}$
  • D
    $n \pi+\frac{\pi}{4}, \frac{\pi}{4}$

Solution

The given system has a non-trivial solution, if $\left|\begin{array}{rrr} \lambda & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ -1 & \sin \alpha & -\cos \alpha \end{array}\right|=0$
By expanding the determinant along ( $C_1$ ), we get $\begin{gathered} \lambda\left(-\cos ^2 \alpha-\sin ^2 \alpha\right)-1(-\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha) \\ -1\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ \lambda=\sin 2 \alpha+\cos 2 \alpha \end{gathered}$ For $\quad \lambda=1, \sin 2 \alpha+\cos 2 \alpha=1$ $\Rightarrow \quad \frac{1}{\sqrt{2}} \sin 2 \alpha+\frac{1}{\sqrt{2}} \cos 2 \alpha=\frac{1}{\sqrt{2}}$ $\Rightarrow \quad \cos \left(2 \alpha-\frac{\pi}{4}\right)=\cos \left(2 n \pi \pm \frac{\pi}{4}\right)$ $\Rightarrow 2 \alpha=2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{4}, n$ being an integer. $\Rightarrow \quad \alpha=n \pi+\frac{\pi}{4}, n \pi$

Hence, the correct answer is option: 'B'.

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