QJEEM18Ma40e5: JEE Main 2018 Mathematics - Class 12 JEEM18Ma40e5 Solution: If the system of linear equations x + ky + 3z = 0 3x + ky - 2z…

System of equations has non-zero solution when determinant of coefficient = 0.

So, in this questions,

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

$\Rightarrow \,\,\,\,$ ( $-$ 3K + 8) $-$ K ( $-$ 9 + 4) + 3(12 $-$ 2K) = 0

$\Rightarrow \,\,\,\,$ $-$ 3K + 8 + 9K $-$ 4K + 36 $-$ 6K = 0

$\Rightarrow \,\,\,\,$ $-$ 4K + 44 = 0

$\Rightarrow \,\,\,\,$ K = 11

Now the equations become

x + 11y + 3z = 0 . . . (1)

3x + 11y $-$ 2z = 0 . . . (2)

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

By adding equation (1) and (3) we get,

3x + 15y = 0

$\Rightarrow \,\,\,\,$ x = $-$ 5y

Putting x = $-$ 5y in equation (1) we get

$-$ 5y + 11y + 3z = 0

$\Rightarrow \,\,\,\,$ 6y + 3z = 0

$\Rightarrow \,\,\,\,$ z = $-$ 2y

$\therefore\,\,\,\,$ ${{xz} \over {{y^2}}}$

$= {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$

$= {{10{y^2}} \over {{y^2}}}$

$= 10$