The set of the 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,
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : contains two elements
The given system of equations is
$x_1-(2-\lambda )-2x_2+x_3=0$
$2x_1-(3+\lambda )x_2+2x_3=0$
$-x_1+2x_2- \lambda x_3= 0 $
Clearly, it is homogenous system of equations and will have non-trivial solutions, if
$\begin{vmatrix}2-\lambda & -2 & 1\\2 & -(3+\lambda) & 2\\-1 & 2 & -\lambda \end{vmatrix}=0$
$⇒ \lambda^3 - \lambda^2 -5\lambda + 3= 0 $
$⇒ \lambda^2( \lambda - 1) + 2 \lambda ( \lambda - 1) - 3 ( \lambda -1) = 0 $
$⇒ ( \lambda -1) ( \lambda^2 + 2 \lambda-3) = 0 $
$⇒( \lambda -1) ( \lambda-1) ( \lambda +3) = 0 $
$⇒ \lambda=1, 1, 3.$
Hence, there are two values of $ \lambda .$