Question

If the system of linear equations

$$\begin{gathered} x+y+3z=0 \\ x+3y+k^{2}z=0 \\ 3x+y+3z=0 \end{gathered}$$

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

$-3$

Explanation for the correct option.

Step 1: Find the value of $k^2$.

As the system of equation has non-zero solutions so $D=0$.

$$\begin{gathered} \left|\begin{array}{ccc}1& 1& 3\\ 1& 3& k^2\\ 3& 1& 3\end{array}\right|=0 \\ \Rightarrow (9-k^2)-(3-3k^2)+3(-8)=0 \\ \Rightarrow 9-k^2-3+3k^2-24=0 \\ \Rightarrow 2k^2-18=0 \\ \Rightarrow k^2=9 \end{gathered}$$

Step 2: Find the value of $x+\frac{y}{z}$

The system of equations for $k^2=9$ is

$$\begin{gathered} x+y+3z=0...\left(i\right) \\ x+3y+9z=0...\left(ii\right) \\ 3x+y+3z=0...\left(iii\right) \end{gathered}$$

Subtract equation $\left(i\right)$ from equation $\left(ii\right)$

$$\begin{gathered} 2y+6z=0 \\ \Rightarrow y=-3z \\ \Rightarrow \frac{y}{z}=-3 \end{gathered}$$

Now substitute $y=-3z$ in equation $\left(iii\right)$

$$\begin{gathered} 3x-3z+3z=0 \\ \Rightarrow 3x=0 \\ \Rightarrow x=0 \end{gathered}$$

Thus the value of $x+\frac{y}{z}$ is $0+\left(-3\right)=-3$.

Hence, the correct option is C.