The set of all values of lambda for which the | vedclass

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Question

$2 x_{1}-2 x_{2}+x_{3}=\lambda x_{1}$

$2 x_{1}-3 x_{2}+2 x_{3}=\lambda x_{2}$

$-x_{1}+2 x_{2}=\lambda x_{3}$

$(2-\lambda) x_{1}-2 x_{2}+x_{3}

Vedclass · Accepted Answer

contains two elements

Vedclass · Answer

$2 x_{1}-2 x_{2}+x_{3}=\lambda x_{1}$

$2 x_{1}-3 x_{2}+2 x_{3}=\lambda x_{2}$

$-x_{1}+2 x_{2}=\lambda x_{3}$

$(2-\lambda) x_{1}-2 x_{2}+x_{3}=0$

$2 x_{1}-(3+\lambda) x_{2}+2 x_{3}=0$

$=-x_{1}+2 x_{2}-\lambda x_{3}=0$

$\angle=0$

$\left|\begin{array}{ccc}{2-\lambda} & {-2} & {1} \ {2} & {-(3+\lambda)} & {2} \ {-1} & {2} & {-\lambda}\end{array}ight|=0$

${R_1} 	o {R_1} + {R_3}$

$\left|\begin{array}{ccc}{1-\lambda} & {0} & {1-\lambda} \ {2} & {-(3+\lambda)} & {2} \ {-1} & {2} & {-\lambda}\end{array}ight|=0$

$=(1-\lambda)[(3+\lambda)(\lambda)-4]+(1-\lambda)[4-(3+\lambda)]=0$

$=(1-\lambda)\left[3 \lambda+\lambda^{2}-4+4-3-\lambdaight]=0$

$=(1-\lambda)\left[\lambda^{2}+2 \lambda-3ight]=0$

$=(1-\lambda)[\lambda(\lambda+3)-1(\lambda+3)]=0$

$=(1-\lambda)(\lambda-1)(\lambda+3)=0$

$\lambda=1,1,3$

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

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