1.

If the system of linear equationsx+ky+3z=03x+ky-2z=02x+4y-3z=0has a non-zero solution (x,y,z) then `(xz)/(y^2)` is equal toA. 30B. -10C. 10D. -30

Answer» Correct Answer - 3
Given system of equations
`x+ky+3z=0`
`3x+ky -2x =0`
`2x+4y -3z =0`
Eliminating from first two equations we get
`2x-5z=0 " or " 2x=5z`
` :. X=5lambda , z= 2lambda`
from equation (iii) we have
`10 lambda +4y-6lambda= 0 " or " y=-lambda`
`:. (xz)/(y^(2)) =((5lambda)(2lambda))/((-lambda)^(2)) =10`


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