If the system of linear equations
$x+y+3 z=0$
$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
has a non-zero solution $(x, y, z)$ for some $k \in \mathbf{R}$, then
$x+\left(\frac{y}{z}\right)$ is equal to :
Correct Option: 1
(1) Since, system of linear equations has non-zero solution
$\therefore \Delta=0$'
$\Rightarrow\left|\begin{array}{ccc}1 & 1 & 3 \\ 1 & 3 & k^{2} \\ 3 & 1 & 3\end{array}\right|=0$
$\Rightarrow 1\left(9-k^{2}\right)-1\left(3-3 k^{2}\right)+3(1-9)=0$
$\Rightarrow 9-k^{2}-3+3 k^{2}-24=0$
$\Rightarrow 2 k^{2}=18 \Rightarrow k^{2}=9, k=\pm 3$
So, equations are
$x+y+3 z=0$ ...(i)
$x+3 y+9 z=0$...(ii)
$3 x+y+3 z=0$...(iii)
Now, from equation (i)-(ii),
$-2 y-6 z=0 \Rightarrow y=-3 z \Rightarrow \frac{y}{z}=-3$ ...(iv)
Now, from equation (i) - (iii),
$-2 x=0 \Rightarrow x=0$
So, $x+\frac{y}{z}=0-3=-3$
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