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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
152:1:3
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B
152:1:3
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C
152:1:3
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D
152:1:3
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Solution

The correct option is D $$ \displaystyle -\frac{15}{2} : 1 : - 3$$
For non trivial solution
$$\Delta = 0$$
$$\therefore \begin{vmatrix}1 & k & 3\\ 3 & k & -2\\ 2 & 3 & -4\end{vmatrix} = 0$$
applying $$R_2 \rightarrow R_2 - 3R_1$$ and $$R_3 \rightarrow R_3 - 2 R_1$$
$$\therefore \begin{vmatrix}1 & k & 3\\ 0 & -2k & -11\\ 0 & 3-2k & -10\end{vmatrix} = 0$$
$$\Rightarrow \begin{vmatrix}-2k & -11 \\3-2k  & -10\end{vmatrix} = 0$$
$$\Rightarrow 20k + 33- 22k = 0$$
$$\therefore k = \dfrac{33}{2}$$
Putting the value of $$k$$ in the given equations. Then equations become
$$\displaystyle x + \dfrac{33}{2} y + 3z = 0$$               ...(i)
$$\displaystyle 3x + \dfrac{33}{2} y - 2z = 0$$              ...(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), $$2x + 3 \lambda + 12 \lambda = 0$$
$$\therefore x = \displaystyle - \frac{15 \lambda}{2}$$
$$\therefore x : y : z = - \displaystyle \frac{15}{2} : 1 : - 3$$

Mathematics

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