Question
Let $$\lambda $$ and $$\alpha $$ be real. The set of all values of $$x$$ for which the system of linear equations
$$\eqalign{
& \lambda x + \left( {\sin \alpha } \right)y + \left( {\cos \alpha } \right)z = 0 \cr
& x + \left( {\cos \alpha } \right)y + \left( {\sin \alpha } \right)z = 0 \cr
& - x + \left( {\sin \alpha } \right) - \left( {\cos \alpha } \right)z = 0 \cr} $$
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 :
$$\left[ { - \sqrt 2 ,\sqrt 2 } \right]$$
Solution :
Since the system has a non-trivial solution, therefore \[\left| {\begin{array}{*{20}{c}}
\lambda &{\sin \alpha }&{\cos \alpha }\\
1&{\cos \alpha }&{\sin \alpha }\\
{ - 1}&{\sin \alpha }&{ - \cos \alpha }
\end{array}} \right| = 0\]
$$\eqalign{
& \Rightarrow \lambda \left( { - {{\cos }^2}\alpha - {{\sin }^2}\alpha } \right) - \left( { - \sin \alpha \cos \alpha - \sin \alpha \cos \alpha } \right) - \left( {{{\sin }^2}\alpha - {{\cos }^2}\alpha } \right) = 0 \cr
& \Rightarrow - \lambda + \sin 2\alpha + \cos 2\alpha = 0 \cr
& \Rightarrow \lambda = \sin 2\alpha + \cos 2\alpha \cr
& \Rightarrow \lambda = \sqrt 2 \cos \left( {2\alpha - \frac{\pi }{4}} \right). \cr
& {\text{Since, }} - 1 \leqslant \cos \left( {2\alpha - \frac{\pi }{4}} \right) \leqslant 1\forall \in R \cr
& \therefore - \sqrt 2 \leqslant \lambda \leqslant \sqrt 2 {\text{ }}\,\,{\text{i}}{\text{.e}}{\text{., }}\lambda \in \left[ { - \sqrt 2 ,\sqrt 2 } \right] \cr} $$