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

The correct option is D exactly three values of $\lambda$

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

$R_3=R_3-R_1$

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

$R_2=R_2-R_1$

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

Expand it

$\Rightarrow -\lambda ({\lambda }^2-1)=0$

$\lambda =0,1,-1$