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  • I have a doubt, kindly clarify. The system of linear equationsx +λy −z = 0λx − y − z = 0x + y − λz = 0has a non-trivial solution for :

The system of linear equations

x +λy −z = 0
λx − y − z = 0
x + y − λz = 0

has a non-trivial solution for :

 

  • Option 1)

    infinitely many values of λ.

     

  • Option 2)

    exactly one value of λ.

     

  • Option 3)

    exactly two values of λ.

     

  • Option 4)

    exactly three values of λ.

     

 

Answers (3)

best_answer

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} 1 & \lambda &-1 \\ \lambda & -1 &-1 \\ 1 & 1 & -\lambda \end{vmatrix}=0

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

\Rightarrow \left ( 1+\lambda \right )\left [ \lambda ^{2}-\lambda \right ]=0

\Rightarrow \lambda = -1,\ 0,\ 1 


Option 1)

infinitely many values of λ.

 

This option is incorrect.

Option 2)

exactly one value of λ.

 

This option is incorrect.

Option 3)

exactly two values of λ.

 

This option is incorrect.

Option 4)

exactly three values of λ.

 

This option is correct.

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