Question

The set of all values of $\lambda$ for which the system of linear equations
$2x_1-2x_2+x_3=\lambda x_1$
$2x_1-3x_2+2x_3=\lambda x_2$
$-x_1+2x_2=\lambda x_3$
has a non-trivial solution

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

Answer 1

Answer: (C)

Given system of linear equations
$2 x_{1}-2 x_{2}+x_{3} =\lambda x_{1} ....$ (i)
$\Rightarrow (2-\lambda) x_{1}-2 x_{2}+x_{3} =0 .....$(ii)
$2 x_{1}-3 x_{2}+2 x_{3} =\lambda x_{2} .....$(iii)
$\Rightarrow 2 x_{1}-(3+\lambda) x_{2}+2 x_{3} =0 $
$-x_{1}+2 x_{2} =\lambda x_{3}$
$\Rightarrow -x_{1}+2 x_{2}-\lambda x_{3} =0$
Since, the system has non-trivial solution.
$\therefore \begin{bmatrix}2-\lambda & -2 & 1 \\2 & -(3+\lambda) & 2 \\-1 & 2 & \lambda\end{bmatrix}=0$
$\Rightarrow \left.(2-\lambda)\left(3 \lambda+\lambda^{2}-4\right)+2(-2 \lambda+2)+1(4-3)-\lambda\right)=0$
$\Rightarrow (2-\lambda)\left(\lambda^{2}+3 \lambda-4\right)+4(1-\lambda)+(1-\lambda)=0 $
$\Rightarrow (2-\lambda)(\lambda+4)(\lambda-1)+5(1-\lambda)=0$
$\Rightarrow (\lambda-1)[(2-\lambda)(\lambda+4)-5]=0 $
$\Rightarrow (\lambda-1)\left(\lambda^{2}+2 \lambda-3\right)=0$
$\Rightarrow (\lambda-1)[(\lambda-1)(\lambda+8)]=0 $
$\Rightarrow(\lambda-1)^{2}(\lambda+3)=0$
$\Rightarrow \lambda=1,1,-3$
Hence, $\lambda$ contains two elements.