The number of values of k for which the linear equations

4x+ky+2z=0

kx+4y+z=0

2x+2y+z=0

possess a non-zero solution is

  • Option 1)

    1

  • Option 2)

    zero

  • Option 3)

    3

  • Option 4)

    2

 

Answers (2)

As we learnt in 

Cramer's rule for solving system of linear equations -

When \Delta =0 and atleast one of   \Delta_{1},\Delta _{2} and \Delta _{3}  is non-zero , system of equations has no solution

- 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}

 

 4x+ky+2z=0

kx+4y+z=0

2x+2y+z=0

\therefore \begin{vmatrix}4&k&2\\ k&4&1 \\ 2&2&1\end{vmatrix}= 0

4\left ( 2 \right )-k\left ( k-2 \right )+2\left ( 2k-8 \right )= 0

= > 8-k^{2}+2k+4k-16=0

= > 6k-k^{2}-8=0

= > k^{2}-6k+8=0

= >\left ( k-4 \right )\left ( k-2 \right )= 0

\therefore k=2\, \, and \, \,k=4

\therefore k=2


Option 1)

1

Incorrect Option

 

Option 2)

zero

Incorrect Option

 

Option 3)

3

Incorrect Option

 

Option 4)

2

Correct Option

 

N neha

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