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
$2x_1-2x_2+x_3=\lambda x_1....$ (i)
$\Rightarrow (2-\lambda )x_1-2x_2+x_3=\mathrm{0.....}$(ii)
$2x_1-3x_2+2x_3=\lambda x_2.....$(iii)
$\Rightarrow 2x_1-(3+\lambda )x_2+2x_3=0$
$-x_1+2x_2=\lambda x_3$
$\Rightarrow -x_1+2x_2-\lambda x_3=0$
Since, the system has non-trivial solution.
$\therefore \left[\begin{array}{ccc}2-\lambda & -2& 1\\ 2& -(3+\lambda )& 2\\ -1& 2& \lambda \end{array}\right]=0$
$\Rightarrow (2-\lambda )\left(3\lambda +{\lambda }^2-4\right)+2(-2\lambda +2)+1(4-3)-\lambda )=0$
$\Rightarrow (2-\lambda )\left({\lambda }^2+3\lambda -4\right)+4(1-\lambda )+(1-\lambda )=0$
$\Rightarrow (2-\lambda )(\lambda +4)(\lambda -1)+5(1-\lambda )=0$
$\Rightarrow (\lambda -1)[(2-\lambda )(\lambda +4)-5]=0$
$\Rightarrow (\lambda -1)\left({\lambda }^2+2\lambda -3\right)=0$
$\Rightarrow (\lambda -1)[(\lambda -1)(\lambda +8)]=0$
$\Rightarrow (\lambda -1{)}^2(\lambda +3)=0$
$\Rightarrow \lambda =1,1,-3$
Hence, $\lambda$ contains two elements.