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 $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k$ is equal to :

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

Answer 1

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

$2x+3y-z=0\dots \left[1\right]$
$x+ky-2z=0\dots \left[2\right]$
$2x-y+z=0\dots \left[3\right]$

Since, the given system of equations has a non-trivial solution
$\therefore \left|\begin{array}{ccc}2& 3& -1\\ 1& k& -2\\ 2& -1& 1\end{array}\right|=0\Rightarrow k=\frac{9}{2}$

From $\left[1\right]$ and $\left[3\right]$, we get
$4x+2y=0\Rightarrow \frac{x}{y}=-\frac{1}{2}$

From $\left[1\right]$ and $\left[3\right]$, we get
$4y-2z=0\Rightarrow \frac{y}{z}=\frac{1}{2}$

From $\left[1\right]$ and $\left[3\right]$, we get
$8x+2z=0\Rightarrow \frac{z}{x}=-4$

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