Question

If the system of equation $2x+3y-z=0,x+ky-2z=0$ and $2x-y+z=0$ has a non trivial solution $(x,y,z)$ then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k$ is equal to:

• A. $\frac{3}{4}$
• B. $-4$
• C. $\frac{1}{2}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{2}$

Given $2x+3y-z=0,x+Ky-2z=0,2x-y+z=0$

for Non-trivial solution

$\left|\begin{array}{ccc}2& 3& -1\\ 1& K& -2\\ 2& -1& 1\end{array}\right|=0$

By expanding along $R_2$

$\Rightarrow -1[3-1]+K[2+2]-(-2)[-2-6]=0$

$\therefore K=\frac{9}{2}$

the given equations become

$2x+3y-z=\mathrm{0......}\left(1\right)$

$x+\frac{9}{2}y-2z=\mathrm{0.......}\left(2\right)$

$2x-y+z=\mathrm{0..........}\left(3\right)$

$\left(1\right)-\left(3\right)\Rightarrow 4y-2z=0$

$2y=z........\left(4\right)$

$\frac{y}{z}=\frac{1}{2}.......\left(5\right)$

put $z$ from equation (4) into (1)

$2x+3y-2y=0$

$2x+y=0$

$\frac{x}{y}=-\frac{1}{2}.....\left(6\right)$

$\left(6\right)\times \left(5\right)\Rightarrow \frac{x}{z}=-\frac{1}{4}\Rightarrow \frac{z}{x}=-4$

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

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