Matrices and Determinants 4 Question 14
15. The system of linear equations $x+\lambda y-z=0 ; \lambda x-y-z=0 ; x+y-\lambda z=0$ has a non-trivial solution for
(2016 Main)
(a) infinitely many values of
(b) exactly one value of $\lambda$
(c) exactly two values of $\lambda$
(d) exactly three values of $\lambda$
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Answer:
Correct Answer: 15. (d)
Solution:
- Given, system of linear equation is $x+\lambda y-z=0 ; \lambda x-y-z=0 ; x+y-\lambda z=0$
Note that, given system will have a non-trivial solution only if determinant of coefficient matrix is zero,
$ \begin{array}{rr} & \left|\begin{array}{ccc} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ \text { i.e. } & & \\ 1 & 1 & -\lambda \end{array}\right|=0 \\ \Rightarrow & 1(\lambda+1)-\lambda\left(-\lambda^{2}+1\right)-1(\lambda+1)=0 \\ \Rightarrow & \lambda+1+\lambda^{3}-\lambda-\lambda-1=0 \\ \Rightarrow & \lambda^{3}-\lambda=0 \Rightarrow \lambda\left(\lambda^{2}-1\right)=0 \\ \Rightarrow & \lambda=0 \text { or } \lambda= \pm 1 \end{array} $
Hence, given system of linear equation has a non-trivial solution for exactly three values of $\lambda$.
