JEE Main | 2018Math miscellaneousHard
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 is equal to :
Options
A.
10
B.
- 30
C.
30
D.
-10
Solution
For non zero solution = 0
k = 11
Now equations
x + 11y + 3z = 0 .....(1)
3x + 11y - 2z = 0 .....(2)
2x + 4y - 3z = 0 .....(3)
on equation (1) + (3) we get 3x + 15y = 0
x = - 5y
Now put x = - 5y in equation (1)
we get - 5y + 11y + 3z = 0
z = - 2y
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