If the system of linear equations
Question:

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 :

1. (1) $-3$

2. (2) 9

3. (3) 3

4. (4) $-9$

Correct Option: 1

Solution:

(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, $x+\frac{y}{z}=0-3=-3$