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. $30$
• C. $-10$
• D. $10$

Answer 1

The correct option is D $10$

$x+ky+3x=0$ ...(i)

$3x+ky-2z=0$ ...(ii)

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

$\left(i\right)-\left(ii\right)\to 3x+x+ky-ky-2z-3z=0$

$2x=5z-\left(4\right)$

$\left(i\right)+\left(iii\right)\to x+2x+ky+4y=0$

$3x=-(k+4)y$

$\frac{x_3}{y^2}=\frac{x.\left(\frac{2}{5}\right)x}{{\left(\frac{-3}{k+4}\right)}^2.x^2}=\frac{2(k+4{)}^2}{5\times 9}$

$=\frac{2(k+4{)}^2}{45}$

Also, for non-zero soln to exist, $|A|=0$

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

$-3k+8-k(-9+4)+3(12-2k)=0$

$-3k+8+9k-4k+36-6k=0$

$-4k+44=0$

$k=11$

$\therefore \frac{x_3}{y^2}=\frac{2}{45}(k+4{)}^2$

$=\frac{2\times 15\times 15}{45}$

$\frac{x_3}{y^2}=10$