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) 3/4
(B) –4
(C) 1/2
(D) –1/4

Answer 1

Answer:

I think option c..............

Answer 2

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 given by:

Given, the set of equations 2x + 3y - z = 0, x + ky - 2z = 0 and 2x - y + z = 0 has a non – trivial solution

⇒ Δ = 0

Therefore, we have,

$\begin{vmatrix}2&3&-1\\ 1&k&-2\\ 2&-1&1\end{vmatrix} = 0$

$2\cdot \det \begin{pmatrix}k&-2\\ -1&1\end{pmatrix}-3\cdot \det \begin{pmatrix}1&-2\\ 2&1\end{pmatrix}-1\cdot \det \begin{pmatrix}1&k\\ 2&-1\end{pmatrix} = 0$

$2\left(k-2\right)-3\cdot \:5-1\cdot \left(-1-2k\right) = 0$

$4k-18 = 0$

∴  k = 9/2

substituting the value of k in given equations, we have,

2x + 3y - z = 0 ..........(1)

x + 9/2y - 2z = 0

⇒ 2x + 9y - 4z = 0 ............(2)

2x - y + z = 0 ..........(3)

equation (1) - (3) gives

2y = z  ................(4)

$\dfrac{y}{z} = \dfrac{1}{2}$...........................A

from  (1) and (4) we get,

$\dfrac{x}{y} = \dfrac{-1}{2}$ ....................B

$\dfrac{z}{x} = \dfrac{-4}{1}$ ..............C

adding A, B, C and k, we get,

$\dfrac{x}{y}+ \dfrac{y}{z}+ \dfrac{z}{x} + k= \dfrac{-1}{2}+ \dfrac{1}{2}+ \dfrac{-4}{1}+ \dfrac{9}{2}$

$\dfrac{x}{y}+ \dfrac{y}{z}+ \dfrac{z}{x} + k= \dfrac{1}{2}$

Therefore option (C) is correct.