Question

Let the system of lineat equations $2x+3y-z=0,2x+ky-3z=0$ and $2x-y+z=0$ have non trivial non trivial solution then $\frac{x}{y}+\frac{y}{z}+zx+k$ will be

• A. $2$
• B. $3$
• C. $1$
• D. $-4$

Answer 1

Answer: (B)

$3$

$\left|\begin{array}{ccc}2& 3& -1\\ 2& k& -3\\ 2& -1& 1\end{array}\right|=0$
$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$\Rightarrow \left|\begin{array}{ccc}2& 3& -1\\ 0& k-3& -2\\ 0& -4& 2\end{array}\right|=0$
$\Rightarrow k=7$
Equation $1\Rightarrow 2\frac{x}{y}+3-\frac{z}{y}=0$ ......(iv)
Equation $3\Rightarrow 2\frac{x}{y}-1+\frac{z}{y}=0$
adding these two equation we get $4\frac{x}{y}+2=0$
$\Rightarrow \frac{x}{y}=-\frac{1}{2}$.....(v)
Putting this value in equation (iv) we get $\frac{z}{y}=2$ ....(vi)
Equation (v) divided by equation (vi) $\Rightarrow \frac{x}{z}=-\frac{1}{4}$
So, $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k=-\frac{1}{2}+\frac{1}{2}-4+7=3$