Question

The system of linear equations

$x+\lambda y-z=0\lambda x-y-z=0x+y-\lambda z=0$

has a non-trivial solution for :

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

Answer 1

The correct option is C

exactly three value of .

$\left|\begin{array}{ccc}1& \lambda & -1\\ \lambda & -1& -1\\ 1& 1& -\lambda \end{array}\right|=0$

$1(\lambda +1)-\lambda (-{\lambda }^2+1)+1(-\lambda -1)=0{\lambda }^3-\lambda +\lambda +1-\lambda -1=0{\lambda }^3-\lambda =0\lambda ({\lambda }^2-1)=0\lambda =0,\lambda =\pm 1$

$\text{Exactly three values of}$ $\lambda$