Question

A non-trivial solution of the system of equations $x+\lambda y+2z=0$, $2x+\lambda z=0$, $2\lambda x-2y+3z=0$ is given by $x:y:z=$

• A. $1:2:-2$
• B. $1:-2:2$
• C. $2:1:2$
• D. $2:1:-2$

Answer 1

Answer: (D)

$\left|\begin{array}{ccc}1& \lambda & 2\\ 2& 0& \lambda \\ 2\lambda & -2& 3\end{array}\right|=0$
$\Rightarrow \left(2\lambda \right)-\lambda {\left(6-2\lambda \right)}^3+2\left(-4\right)=0$
$\Rightarrow {\lambda }^3-2\lambda -4=0\Rightarrow \left(\lambda -2\right)\left({\lambda }^2+2\lambda +2\right)=0$
Taking $\lambda =2$, the system becomes
$x+2y+2z=0$
$2x+0y+2z=0$
$4x-2y+3z=0$
$\therefore \frac{x}{4}=\frac{y}{2}=\frac{z}{-4}$
$\Rightarrow x:y:z=2:1:-2.$