Question

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

• A. $-9$
• B. $9$
• C. $-3$
• D. $3$

Answer 1

The correct option is C $-3$

Given:
$x+y+3z=0\cdots \left(i\right)$
$x+3y+k^{2}z=0\cdots \left(ii\right)$
$3x+y+3z=0\cdots \left(iii\right)$
$\left|\begin{array}{ccc}1& 1& 3\\ 1& 3& k^2\\ 3& 1& 3\end{array}\right|=0$
$\Rightarrow (9-k^2)-(3-3k^2)+3(-8)=0$
$\Rightarrow 9-k^2-3+3k^2-24=0$
$\Rightarrow 2k^2-18=0$
$\Rightarrow k^2=9$
$\Rightarrow k=3,-3$

Solving $\left(i\right)$ and $\left(ii\right),$ we get
$\Rightarrow 2y+6z=0$
$\Rightarrow y=-3z$
$\Rightarrow \frac{y}{z}=-3$
Put in equation $\left(i\right),$ we get
$x=0$
So, $x+\frac{y}{z}=-3$