Question

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 :

• A. infinitely many values of $\lambda$
• B. exactly one value of $\lambda$
• C. exactly two values of $\lambda$
• D. exactly three values of $\lambda$

Answer 1

Answer: (D)

$x+\lambda y-z=0$
$\lambda x-y-z=0$
$x+y-\lambda z=0$
For non-trivial solution $\Rightarrow \Delta =0$
$\Rightarrow \left|\begin{array}{ccc}1& \lambda & -1\\ \lambda & -1& -1\\ 1& 1& -\lambda \end{array}\right|=0$
$\lambda +1-\lambda \{-{\lambda }^2+1\}-(\lambda +1)=0$
$\lambda ({\lambda }^2-1)=0$
$\lambda =0,\pm 1$
Exactly $3$ values of $\lambda$.