Question

If the system of equations, `2x + 3y-z = 0, x + ky -2z = 0 " and " 2x-y+z = 0` has a non-trivial solution (x, y, z), then `(x)/(y) + (y)/(z) + (z)/(x) + k` is equal to
A. `-4`
B. `(1)/(2)`
C. `-(1)/(4)`
D. `(3)/(4)`

Answer 1

Given system of linear equations
2x+ 3y-z = 0,
x + ky-2z = 0
and 2x-y+z = 0 has a non-trivial solution (x, y, z).
`therefore Delta = 0 rArr |{:(2, 3, -1), (1, k, -2), (2, -1, 1):}| = 0`
`2(k-2)-3 (1+4)-1(-1-2k) = 0`
`rArr 2k-4-15 +1+2k = 0`
`rArr 4k = 18 rArr k = (9)/(2)`
So, system of linear equations is
`2x +3y-z = 0 " "...(i)`
`2x+ 9y-4z = 0 " "...(ii)`
`"and " 2x-y+z = 0 " "...(iii)`
From Eqs. (i) and (ii), we get
`6y-3z = 0, (y)/(z) = (1)/(2)`
From Eqs. (i) and (iii), we get
`4x + 2y = 0 rArr (x)/(y) = -(1)/(2)`
`"So, "(x)/(2) = (x)/(y) xx (y)/(z) = -(1)/(4) rArr (z)/(x) = -4 " "[because (y)/(z) = (1)/(2) " and " (x)/(y) = -(1)/(2)]`
`therefore (x)/(y) + (y)/(z) + (z)/(x) + k = -(1)/(2) + (1)/(2) -4 + (9)/(2) = (1)/(2)`