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,

  • Option 1)

    is an empty set.

  • Option 2)

    is a singleton.

  • Option 3)

    contains two elements.

  • Option 4)

    contains more than two elements.

 

Answers (2)

As we learnt in 

Cramer's rule for solving system of linear equations -

When \Delta =0  and \Delta _{1}=\Delta _{2}=\Delta _{3}=0 ,

then  the system of equations has infinite solutions.

- wherein

a_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}

a_{3}x+b_{3}y+c_{3}z=d_{3}

and 

\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}

\Delta _{1},\Delta _{2},\Delta _{3} are obtained by replacing column 1,2,3 of \Delta by \left ( d_{1},d_{2},d_{3} \right )  column

 

\begin{vmatrix} 2-\lambda & -2 & 1\\ 2 & -3-\lambda & 2\\ -1 & 2 & -\lambda \end{vmatrix}=0

\therefore \left ( 2-\lambda \right )\left ( \lambda ^{2}+3\lambda -4 \right )+4\left ( 1-\lambda \right )+\left ( 1-\lambda \right )=0

\therefore \left ( \lambda -1 \right )\left ( \lambda ^{2}+2\lambda -3 \right )=0

\left ( \lambda +3 \right )\left ( \lambda -1 \right )^{2}=0

\therefore \lambda =1\:and\:\lambda =-3 


Option 1)

is an empty set.

This option is incorrect.

Option 2)

is a singleton.

This option is incorrect.

Option 3)

contains two elements.

This option is correct.

Option 4)

contains more than two elements.

This option is incorrect.

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