Question

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

• A. $10$
• B. $-30$
• C. $30$
• D. $-10$

Answer 1

The correct option is A $10$

Solution:-

The system of linear equations given are

$x+ky+3z=0$

$3x+ky–2z=0$

$2x+4y–3z=0$

For these equations to have a non-zero solution

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

$\Rightarrow 1(-3k-(-8))-k(-9-\left(4\right))+3(12-2k)=0$

$\Rightarrow -3k+8+5k+36-6k=0$

$\Rightarrow 44-4k=0$

$\Rightarrow k=11$

The equations will become

$x+11y+3z=0...............\left(1\right)$

$3x+11y–2z=0...............\left(2\right)$

$2x+4y–3z=0...............\left(3\right)$

Adding ${eq}^n\left(1\right)\&\left(3\right)$, we get

$(x+11y+3z)+(2x+4y-3z)=0$

$3x+15y=0$

$\Rightarrow x+5y=0$

$\Rightarrow x=-5y...............\left(4\right)$

Putting the value of x in ${eq}^n\left(3\right)$, we have

$2\times (-5y)+4y-3z=0$

$\Rightarrow -10y+4y-3z=0$

$\Rightarrow -6y-3z=0$

$\Rightarrow 2y+z=0$

$\Rightarrow z=-2y$

$\therefore \frac{xz}{y^2}=\frac{(-5y)(-2y)}{{\left(y\right)}^2}=\frac{10y^2}{y^2}=10$

Hence, the correct answer is 10.