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 + (y/z)` is equal to :

Options

A. `9`
B. `-3`
C. `-9`
D. `3`

Answer

To solve the given system of linear equations and find the value of \( x + \frac{y}{z} \), we will follow these steps:

### Step 1: Write the system of equations
The system of equations is given as:
1. \( x + y + 3z = 0 \) (Equation 1)
2. \( x + 3y + k^2 z = 0 \) (Equation 2)
3. \( 3x + y + 3z = 0 \) (Equation 3)

### Step 2: Formulate the coefficient matrix and find the determinant
To determine if there is a non-zero solution, we need to find the determinant of the coefficient matrix:
\[ \begin{vmatrix} 1 & 1 & 3 \\ 1 & 3 & k^2 \\ 3 & 1 & 3 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:
\[ D = 1 \cdot \begin{vmatrix} 3 & k^2 \\ 1 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & k^2 \\ 3 & 3 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 3 & k^2 \\ 1 & 3 \end{vmatrix} = 9 - k^2 \)
2. \( \begin{vmatrix} 1 & k^2 \\ 3 & 3 \end{vmatrix} = 3 - 3k^2 \)
3. \( \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} = 1 - 9 = -8 \)

Putting it all together:
\[ D = 1(9 - k^2) - 1(3 - 3k^2) + 3(-8) \]
\[ D = 9 - k^2 - 3 + 3k^2 - 24 \]
\[ D = 2k^2 - 18 \]

### Step 4: Set the determinant to zero for non-zero solutions
For the system to have a non-zero solution, the determinant must be zero:
\[ 2k^2 - 18 = 0 \]
\[ 2k^2 = 18 \implies k^2 = 9 \implies k = 3 \text{ or } k = -3 \]

### Step 5: Substitute \( k^2 \) back into the equations
Substituting \( k^2 = 9 \) into Equation 2, we get:
\[ x + 3y + 9z = 0 \quad \text{(Equation 2')} \]

Now we have the modified system:
1. \( x + y + 3z = 0 \)
2. \( x + 3y + 9z = 0 \)
3. \( 3x + y + 3z = 0 \)

### Step 6: Solve the equations
From Equation 1:
\[ x + y + 3z = 0 \implies x = -y - 3z \]

Substituting \( x \) into Equation 2':
\[ (-y - 3z) + 3y + 9z = 0 \]
\[ 2y + 6z = 0 \implies y = -3z \]

Now substituting \( y \) back into the expression for \( x \):
\[ x = -(-3z) - 3z = 3z - 3z = 0 \]

### Step 7: Find \( x + \frac{y}{z} \)
Now we have:
- \( x = 0 \)
- \( y = -3z \)

Thus:
\[ x + \frac{y}{z} = 0 + \frac{-3z}{z} = -3 \]

### Final Answer
The value of \( x + \frac{y}{z} \) is \( \boxed{-3} \).