Q49 / 90 · JEE Main 2018 (08 Apr) · Mathematics

If the system of linear equations
$x + k y + 3 z = 0$
$3 x + k y - 2 z = 0$
$2 x + 4 y - 3 z = 0$
has a non-zero solution $(x, y, z),$ then $\frac{x z}{y^{2}}$ is equal to:

Solution

|\begin{matrix} 1 & k & 3 \\ 3 & k & - 2 \\ 2 & 4 & - 3 \end{matrix}| = 0

\Rightarrow (- 3 k + 8) - 3 (- 3 k - 12) + 2 (- 5 k) = 0

\Rightarrow - 4 k + 44 = 0

\Rightarrow

k = 11

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

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

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

On solving equations

(1) & (2)

, we get

2 x - 5 z = 0

\Rightarrow x = \frac{5 z}{2}

Put it in equation

(3)

, we get

2 z + 4 y = 0

\Rightarrow z = - 2 y

∴ \frac{x z}{y^{2}} = \frac{(- 2 y) \times (- 5 y)}{y^{2}} = 10

Hence, the correct answer is option: 'C'.