if the system of equations `x+y+z=0`, `x+3y+k^2z=0` and `x+2y+z=0` have a non zero solution then value of `y+(x/z)` is

Options

A. 1
B. 0
C. -1
D. 2

Answer

To solve the given system of equations for the condition of having a non-zero solution, we will follow these steps:

### Step 1: Set up the equations
We have the following equations:
1. \( x + y + z = 0 \) (Equation 1)
2. \( x + 3y + k^2z = 0 \) (Equation 2)
3. \( x + 2y + z = 0 \) (Equation 3)

### Step 2: Form the coefficient matrix and set the determinant to zero
For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero. The coefficient matrix is:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 3 & k^2 \\ 1 & 2 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
We will calculate the determinant \( D \):

\[ D = 1 \cdot \begin{vmatrix} 3 & k^2 \\ 2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & k^2 \\ 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 3 & k^2 \\ 2 & 1 \end{vmatrix} = 3 \cdot 1 - k^2 \cdot 2 = 3 - 2k^2 \)
2. \( \begin{vmatrix} 1 & k^2 \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - k^2 \cdot 1 = 1 - k^2 \)
3. \( \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} = 1 \cdot 2 - 3 \cdot 1 = 2 - 3 = -1 \)

Putting it all together:

\[ D = 1(3 - 2k^2) - 1(1 - k^2) + 1(-1) \]
\[ D = 3 - 2k^2 - 1 + k^2 - 1 \]
\[ D = 1 - k^2 \]

### Step 4: Set the determinant to zero
For the system to have a non-zero solution, we set the determinant to zero:

\[ 1 - k^2 = 0 \]

This gives us:

\[ k^2 = 1 \implies k = \pm 1 \]

### Step 5: Substitute \( k \) back into the equations
Now we can consider both cases for \( k \):

1. **Case \( k = 1 \)**:
- The equations become:
1. \( x + y + z = 0 \)
2. \( x + 3y + z = 0 \)
3. \( x + 2y + z = 0 \)

2. **Case \( k = -1 \)**:
- The equations remain the same as above.

### Step 6: Solve the equations
Let's solve the equations for \( k = 1 \):

From Equation 1:
\[ z = -x - y \]

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

Now substituting \( y = 0 \) back into Equation 1:
\[ x + 0 + z = 0 \implies z = -x \]

### Step 7: Calculate \( y + \frac{x}{z} \)
Now we can find \( y + \frac{x}{z} \):
\[ y + \frac{x}{z} = 0 + \frac{x}{-x} = -1 \]

### Final Answer
Thus, the value of \( y + \frac{x}{z} \) is \( -1 \).

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