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If the system of linear equations  x+y+3z=0 x+3y+k^2z=0 3x+y+3z=0 has non zero solution (x,y,z)\: \: for\, \, \: \: some \: \: k\in R then x+\left ( \frac{y}{z} \right )=?   
Option: 1 -9
Option: 2 9
Option: 3 3
Option: 4 -3

Answers (1)

best_answer

For non-zero solutions to exist,

\\\begin{vmatrix} 1 &1 &3 \\ 1& 3 &k^2 \\ 3& 1 &3 \end{vmatrix}=0\\\Rightarrow 1(9-k^2)-1(3-3k^2)+3(1-9)=0\\k^2=9\\

Now the system of equation become

x + y + 3z = 0  

x + 3y + 9z = 0

3x + y + 3z = 0

we, get x = 0

3y = -9z

y/z = -3

x+ (y/z) = -3

Posted by

himanshu.meshram

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