Question

The system of linear equations $x+\lambda y-z-y-z=0,\lambda x-y-z=0,x+y-\lambda z=0$ has a non-trivial solutions for:

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

Answer 1

Answer: (A)

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.

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

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

$⟹\lambda +1+{\lambda }^3-2\lambda -1=0$

$⟹{\lambda }^3-\lambda =0⟹(\lambda -1{)}^2=0$

$⟹\lambda =0$ and $\lambda =\pm 1$

Hence, given system of linear equation has a non - trivial solution for exactly three values of

$\lambda$