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

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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