Question

If the system of equations

$$\begin{gathered} x+y+z=2 \\ 2x+4y–z=6 \\ 3x+2y+\lambda z=\mu \end{gathered}$$

has infinitely many solutions, then

• A. $\lambda -2\mu =-5$
• B. $2\lambda +\mu =14$
• C. $\lambda +2\mu =14$
• D. $2\lambda -\mu =5$

Answer 1

The correct option is B

$2\lambda +\mu =14$

Explanation for the correct option.

Step 1. Find the value of $\lambda$.

The given system has infinitely many solutions so $D=D_1=D_2=D_3=0$.

Now for the system $$\begin{gathered} x+y+z=2 \\ 2x+4y–z=6 \\ 3x+2y+\lambda z=\mu \end{gathered}$$ taking $D=0$ the equation can be formed as:

$$\begin{gathered} \left|\begin{array}{ccc}1& 1& 1\\ 2& 4& -1\\ 3& 2& \lambda \end{array}\right|=0 \\ \Rightarrow 1\left(4\lambda +2\right)-1\left(2\lambda +3\right)+1\left(4-12\right)=0 \\ \Rightarrow 4\lambda +2-2\lambda -3-8=0 \\ \Rightarrow 2\lambda =9 \\ \Rightarrow \lambda =\frac{9}{2} \end{gathered}$$

Step 2. Find the value of $\mu$.

Again solve for $D_3=0$.

$$\begin{gathered} \left|\begin{array}{ccc}1& 1& 2\\ 2& 4& 6\\ 3& 2& \mu \end{array}\right|=0 \\ \Rightarrow 1\left(4\mu -12\right)-1\left(2\mu -18\right)+2\left(4-12\right)=0 \\ \Rightarrow 4\mu -12-2\mu +18-16=0 \\ \Rightarrow 2\mu =10 \\ \Rightarrow \mu =5 \end{gathered}$$

Step 3. Find the relation.

For $\lambda =\frac{9}{2}$ and $\mu =5$, the value of $2\lambda +\mu$ is:

$\begin{array}{rcl}2\lambda +\mu & =& 2\times \frac{9}{2}+5\\ & =& 9+5\\ & =& 14\end{array}$

Hence, the correct option is B.