Determinants Matrices Question 12
Question: Let $ \lambda $ and $ \alpha $ be real. The set of all values of x 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 )-(\cos \alpha )z=0 $ has a non-trivial solution, is
Options:
A) $ [ 0,\sqrt{2} ] $
B) $ [ -\sqrt{2},0 ] $
C) $ [ -\sqrt{2},\sqrt{2} ] $
D) None of these
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Answer:
Correct Answer: C
Solution:
- [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 \lambda (-{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha ) $
$ -(-\sin \alpha \cos \alpha -\sin \alpha \cos \alpha ) $
$ -({{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha )=0 $
$ \Rightarrow -\lambda +\sin 2\alpha +\cos 2\alpha =0 $
$ \Rightarrow \lambda =\sin 2\alpha +\cos 2\alpha $
$ \Rightarrow \lambda =\sqrt{2}\cos ( 2\alpha -\frac{\pi }{4} ). $ Since $ -1\le \cos ( 2\alpha -\frac{\pi }{4} )\le 1\forall \in R $
$ \therefore -\sqrt{2}\le \lambda \le \sqrt{2} $ i.e. $ \lambda \in [ -\sqrt{2},\sqrt{2} ] $
BETA
