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  • I have a doubt, kindly clarify. The number of real values of λ for which the system of linear equations2x+4y−λz=04x+λy+2z=0λx+2y+2z=0has infinitely many solutions, is :

 The number of real values of λ for which the system of linear equations

2x+4y−λz=0

4x+λy+2z=0

λx+2y+2z=0

has infinitely many solutions, is :

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    3

 

Answers (2)

best_answer

As we learnt in 

By using the concept of

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 & 4 & -\lambda \\ 4 & \lambda &2 \\ \lambda & 2 & 2 \end{vmatrix} =0

\Rightarrow 2\begin{vmatrix} \lambda & 2 \\ 2 & 2 \end{vmatrix} -4\begin{vmatrix} 4 &2 \\ \lambda & 2 \end{vmatrix} -\lambda \begin{vmatrix} 4 & \lambda \\ \lambda & 2 \end{vmatrix} =0

\Rightarrow 2(2\lambda-4)-4(8-2\lambda)-\lambda(8-\lambda^{2})=0

\Rightarrow 4\lambda - 8 - 32 +8\lambda - 8\lambda+\lambda^{3}=0

\Rightarrow \lambda^{3}+4\lambda - 40 =0

It will give only one real value of \lambda 

 

 


Option 1)

0

Incorrect

Option 2)

1

Incorrect

Option 3)

2

Incorrect

Option 4)

3

Correct

Posted by

prateek

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