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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.

Posted by

Sabhrant Ambastha

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