JEE Main On 08 April 2018 Question 6
Question: If the system of linear equations
$ x+ky+3z=0 $ $ 3x+ky-2z=0 $ $ 2x+4y-3z=0 $ Has a non-zero solution (x, y, z), then $ \frac{xz}{y^{2}} $ is equal to: [JEE Main Online 08-04-2018]
Options:
A) $ -30 $
B) $ 30 $
C) $ -10 $
D) $ 10 $
Show Answer
Answer:
Correct Answer: D
Solution:
For non-zero solution,
$ \therefore x+11y=-3z $ ?(1) $ 3x+11y=2z $ ?(2) $ -2x=-5z $ ?(1)-(2)
$ \Rightarrow x=\frac{5}{2}z $
$ \therefore 11y=-3z-\frac{5}{2}z=-\frac{11}{2}z\Rightarrow y=-\frac{z}{2} $
$ \therefore \frac{xz}{y^{2}}=\frac{\frac{5}{2}z.z}{\frac{z^{2}}{4}}=10 $
