System of linear equations

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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

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Solution

$A h o m o g e n e o u s s y s t e m o f e q u a t i o n s h a s n o n - z e r o s o l u t i o n i f Δ = 0$
$\Rightarrow |\begin{matrix} 1 & k & 3 \\ 3 & k & - 2 \\ 2 & 4 & - 3 \end{matrix}| = 0 \Rightarrow 1 (- 3 k + 8) - k (- 9 + 4) + 3 (12 - 2 k) = 0 \Rightarrow 4 k = 44$
   $\Rightarrow K = 11$
$∴ T h e g i v e n e q u a t i o n s b e c o m e$
$x + 11 y + 3 z = 0 \to (1)$
$3 x + 11 y - 2 z = 0 \to (2) 2 x + 4 y - 3 z = 0 \to (3)$
$(1) + (3) \Rightarrow 3 x + 15 y = 0 \Rightarrow x = - 5 y$
Put $x = - 5 y$ in equation (1),
Le get, $- 5 y + 11 y + 3 z = 0 \Rightarrow Z = - 2 y$
$∴ \frac{x z}{y^{2}} = \frac{- 5 y \times - 2 y}{y^{2}} = 10$

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System of linear equations

Remember concepts with our Masterclasses.

If the system of linear equationsx+ky+3z=0, 3x+ky−2z=0,2x+4y−3z=0 has a non – zero solution x,y,zThen x zy2is equal to

Ready to Test Your Skills?

free study material

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