JEE Main 12 Jan 2019 Evening Question 13
Question: The set of all values of $ \lambda $ for which the system of linear equations
$ x-2y-2z=\lambda x $ $ x+2y+z=\lambda y $ $ -x-y=\lambda z $ has a non-trivial solution [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) is an empty set
B) contains exactly two elements
C) is a singleton
D) contains more than two elements
Show Answer
Answer:
Correct Answer: C
Solution:
- The given system of linear equations is $ (1-\lambda )x-2y-2z=0 $ $ x+(2-\lambda )y+z=0 $ $ -x-y\lambda z=0 $ Since, it has a non-trivial solution.
$ \therefore $ $ \begin{vmatrix} 1-\lambda & -2 & -2 \\ 1 & 2-\lambda & 1 \\ -1 & -1 & -\lambda \\ \end{vmatrix} =0 $
$ \Rightarrow $ $ (1-\lambda )[(2-\lambda )(-\lambda )+1]+2(-\lambda +1) $ $ -2(-1+2-\lambda )=0 $
$ \Rightarrow $ $ (1-\lambda )[-2\lambda +{{\lambda }^{2}}+1+2-2]=0 $
$ \Rightarrow $ $ (1-\lambda ){{(1-\lambda )}^{2}}=0\Rightarrow {{(1-\lambda )}^{3}}=0\Rightarrow \lambda =1 $
