Question

If the system of equations $x+ky+3z=0,3x+ky-2z=0,2x+3y-4z=0$ has non trivial solution, then $\frac{xy}{z^2}=.....$

• A. $\frac{-15}{2}$
• B. $-\frac{5}{6}$
• C. $\frac{6}{5}$
• D. $-\frac{6}{5}$

Answer 1

The correct option is A $\frac{-15}{2}$

$x+ky+3z=0$

$3x+ky-2z=0$

$2x+3y-4z=0$ has non trivial solution then

$\left|\begin{array}{ccc}1& k& 3\\ 3& k& -2\\ 2& 3& 4\end{array}\right|=0$

$4k+6-k(12+4)+3(9-2k)=0$

$4k+6-12k-4k+27-6k=0$

$18k=33$

$k=\frac{11}{6}$

Equations are $6x+11y+18z=0$

$18x+11y-12z=0$

$2x+3y-4z=0$

on solving we get

$x=-\frac{5}{2}$; $y=3$; $z=-1$

$\frac{xy}{z^2}=\frac{-15}{2{(-1)}^2}=\frac{-15}{2}$