Question

the set of all values of λ for which the system lf linear solutions

2x₁ - 2x₂ + x₃ = λx₁
2x₁ - 3x₂ + 2x₃ = λx₂
-x₁ + 2x₂ = λx₃

has a non trivial solution.

(1) is a singleton
(2) contains two elements
(3) contains two or more elements
(4) is an empty set (∅)​

Answer 1

Solution:-

$\boldsymbol{\sf x_1(2-\lambda)-2x_2+x_3=0}$

$\boldsymbol{\sf2x_1+x_2(-\lambda-3)+2x_3=0}$

$\boldsymbol{\sf -x_1+2x_2-\lambda x_3 = 0}$

$\boldsymbol{\sf \left| \begin{array}{ccc} \sf 2-\lambda & \sf -2 &\sf 1\\\sf 2 & \sf -\lambda-3 & \sf 2\\\sf -1&\sf 2&\sf -\lambda\end{array}\right|}$

$\boldsymbol{\sf (2-\lambda)(\lambda^2 + 3\lambda -4) + 2(-2\lambda +2)+(4-\lambda -3) = 0}$

$\boldsymbol{\sf \implies -\lambda^3 -\lambda^2 +5\lambda -3=0}$

$\boldsymbol{\sf \implies \lambda^3 +\lambda^2 -5\lambda +3=0}$

$\boldsymbol{\sf \lambda^3 -\lambda^2 +2\lambda^2 -2\lambda -3\lambda + 3 = 0}$

$\boldsymbol{\sf \lambda^2(\lambda-1)+(2\lambda(\lambda -1) - 3(\lambda -1) = 0}$

$\boldsymbol{\sf (\lambda -1)(\lambda^2 +2\lambda -3) = 0}$

$\boldsymbol{\sf (\lambda-1)(\lambda+3)(\lambda-1) = 0}$

$\boxed{\boldsymbol{\sf \implies \lambda = 1,1,-3}}$

as it contains two elements, therefore second option is correct.