Question

The system of linear equations
$x+\lambda y-z=0;\lambda x-y-z=1;x+y-\lambda z=0$ 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 $\lambda$

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,
i.e. $\left|\begin{array}{ccc}1& \lambda & -1\\ \lambda & -1& -1\\ 1& 1& -\lambda \end{array}\right|=0$
$\Rightarrow 1(\lambda +1)-\lambda (-{\lambda }^2+1)-1(\lambda +1)=0\Rightarrow \lambda +1+{\lambda }^3-\lambda -\lambda -1=0\Rightarrow {\lambda }^3-\lambda =0\Rightarrow ({\lambda }^2-1)=0\Rightarrow \lambda =0or\lambda =\pm 1$
Hence, given system of linear equation has a non - trivial solution for exactly three values of $\lambda$