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)

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