Question

Let $S$ be the set of all $\lambda \in R$ for which the system of linear equations

$$\begin{gathered} 2x-y+2z=0 \\ x-2y+\lambda z=-4 \\ x+\lambda y+z=4 \end{gathered}$$

has no solution. Then the set $S$ is.

• A. is an empty set
• B. is a singleton
• C. contains more than two elements
• D. contains exactly two elements

Answer 1

The correct option is D

contains exactly two elements

Explanation of correct answer :

Determining consistency of linear equations:

If three equations doesn't hold a solution , then their determinant has to be zero, $△=0$.

$$\begin{gathered} \left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}2& -1& 2\\ 1& -2& \lambda \\ 1& \lambda & 1\end{array}\right|=0 \end{gathered}$$

Now, simplify it.

$$\begin{gathered} \Rightarrow 2\left(-2-{\lambda }^2\right)+1(1-\lambda )+2(\lambda +2)=0 \\ \Rightarrow -4-2{\lambda }^2+1-\lambda +2\lambda +4=0 \\ \Rightarrow -2{\lambda }^2+\lambda +1=0 \\ \Rightarrow (\lambda -1)\left(\lambda +\frac{1}{2}\right)=0 \\ \Rightarrow \lambda =1\text{or}\frac{-1}{2} \end{gathered}$$

For two values of $\lambda$ , equation has no solutions.

Thus, the correct answer is Option (D).