Question

If the following system of equations possess a non-trivial solution over the set of rationals
$x+ky+3z=0$
$3x+ky-2z=0$
$2x+3y-4z=0$,
then x,y,z are in the ratio

• A. $\frac{15}{2}:1:3$
• B. $\frac{15}{2}:-1:-3$
• C. $\frac{15}{2}:1:-3$
• D. $-\frac{15}{2}:1:-3$

Answer 1

The correct option is D $-\frac{15}{2}:1:-3$

For non trivial solution

$\Delta =0$

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

applying $R_2\to R_2-3R_1$ and $R_3\to R_3-2R_1$

$\therefore \left|\begin{array}{ccc}1& k& 3\\ 0& -2k& -11\\ 0& 3-2k& -10\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{cc}-2k& -11\\ 3-2k& -10\end{array}\right|=0$

$\Rightarrow 20k+33-22k=0$

$\therefore k=\frac{33}{2}$

Putting the value of $k$ in the given equations. Then equations become

$x+\frac{33}{2}y+3z=0$ ...(i)

$3x+\frac{33}{2}y-2z=0$ ...(ii)

$2x+3y-4z=0$ .....(iii)

Multiply (i) by 3 and subtract from (ii) then we get

$-33y-11z=0$

or $z=-3y$ ...(iv)

again multiply (i) by 2 and subtract from (iii) then we get

$-30y-10z=0$

$\therefore z=-3y$ ....(v)

Now let $y=\lambda ,$

$\therefore z=-3\lambda$

from (iii), $2x+3\lambda +12\lambda =0$

$\therefore x=-\frac{15\lambda }{2}$

$\therefore x:y:z=-\frac{15}{2}:1:-3$