Question

If the system 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. $-30$
• B. $10$
• C. $-10$
• D. $30$

Answer 1

The correct option is B $10$

For a non-zero solution $D=0$
$\left|\begin{array}{ccc}1& k& 3\\ 3& k& -2\\ 2& 4& -3\end{array}\right|=0$
$\Rightarrow 1(-3k+8)-k(-9+4)+3(12-2k)=0$
$\Rightarrow -3k+8+5k+36-6k=0$
$\Rightarrow -4k+44=0$
$k=11$
Hence equations are $x+11y+3z=0$
$3x+11y-2z=0$
and $2x+4y-3z=0$
Let $z=t$
then we get $x=\frac{5}{2}t$ and $y=-\frac{t}{2}$
thus, $\frac{xz}{y^2}=\frac{\left(\frac{5}{2}t\right)\left(t\right)}{\frac{t^2}{4}}=10$