Question

Let the system of linear equations$4x+\lambda y+2z=0$, $2x-y+z=0$ and$\mu x+2y+3z=0$, $\lambda$ $\mu \in R$ 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

$\mu =6,\lambda \in R$

Explanation for the correct option:

The given system of linear equations is $4x+\lambda y+2z=0$, $2x-y+z=0$ and$\mu x+2y+3z=0$, $\lambda$, $\mu \in R$.

We know that for non-trivial solution determinant is zero, $\left|\begin{array}{ccc}4& \lambda & 2\\ 2& -1& 1\\ \mu & 2& 3\end{array}\right|=0$

The determinant is $0$, this gives:

$$$$\begin{gathered} \Rightarrow 4\left(-3-2\right)-\lambda \left(6-\mu \right)+2\left(4+\mu \right)=0 \\ \Rightarrow -20-6\lambda +\lambda \mu +8+2\mu =0 \\ \Rightarrow 2\mu -6\lambda +\lambda \mu -12=0 \end{gathered}$$

So when $\mu =6$

$2\left(6\right)-6\lambda +6\lambda -12=0$, this equation is satisfied

Hence, the correct answer is option (A).