Question

Let the system of linear equations
$4x+\lambda y+2z=0$
$2x-y+z=0$
$\mu +2y+3z=0,\lambda ,\mu \in R$
has a non-trivial solution. Then which of the following is true?

• A. $\mu =6,\lambda \in R$
• B. $\lambda =2,\mu \in R$
• C. $\lambda =3,\mu \in R$
• D. $\mu =-6,\lambda \in R$

Answer 1

The correct option is A $\mu =6,\lambda \in R$

For non-trivial solution
$\Delta =0\Rightarrow \left|\begin{array}{ccc}4& \lambda & 2\\ 2& -1& 1\\ \mu & 2& 3\end{array}\right|=0\Rightarrow 4(-5)-\lambda (6-\mu )+2(4+\mu )=0\Rightarrow -12-6\lambda +\lambda \mu +2\mu =0\Rightarrow -6(\lambda +2)+\mu (\lambda +2)=0\Rightarrow (\lambda +2)(\mu -6)=0\Rightarrow \lambda =-2\text{or}\mu =6$
When $\lambda =-2$, first and second equation becomes same, so
$\mu =6,\lambda \in R$