If the system of equations, `2x + 3y-z = 0, x + ky -2z = 0 " and " 2x-y+z = 0` has a non-trivial solution (x, y, z), then `(x)/(y) + (y)/(z) + (z)/(x) + k` is equal to

Options

A. `-4`
B. `1/2`
C. `-1/4`
D. `3/4`

Answer

To solve the problem step by step, we will analyze the given system of equations and find the required expression.

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

### Step 2: Form the coefficient matrix
The coefficient matrix \( A \) for the system of equations is:
\[ A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & k & -2 \\ 2 & -1 & 1 \end{bmatrix} \]

### Step 3: Set the determinant of the matrix to zero for non-trivial solutions
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:
\[ \text{det}(A) = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant using expansion along the second row:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} - k \cdot \begin{vmatrix} 2 & -1 \\ 2 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 3 \\ 2 & -1 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 3 & -1 \\ -1 & 1 \end{vmatrix} = (3)(1) - (-1)(-1) = 3 - 1 = 2 \)

2. \( \begin{vmatrix} 2 & -1 \\ 2 & 1 \end{vmatrix} = (2)(1) - (-1)(2) = 2 + 2 = 4 \)

3. \( \begin{vmatrix} 2 & 3 \\ 2 & -1 \end{vmatrix} = (2)(-1) - (3)(2) = -2 - 6 = -8 \)

Putting these values back into the determinant:
\[ \text{det}(A) = 1 \cdot 2 - k \cdot 4 - 2 \cdot (-8) = 2 - 4k + 16 = 0 \]

### Step 5: Solve for \( k \)
Setting the determinant equal to zero:
\[ 2 - 4k + 16 = 0 \implies 18 = 4k \implies k = \frac{18}{4} = \frac{9}{2} \]

### Step 6: Substitute \( k \) back into the equations
Now we substitute \( k = \frac{9}{2} \) back into Equation 2:
\[ x + \frac{9}{2}y - 2z = 0 \quad \text{(Equation 2)} \]

### Step 7: Simplify the equations
Now we have:
1. \( 2x + 3y - z = 0 \)
2. \( x + \frac{9}{2}y - 2z = 0 \)
3. \( 2x - y + z = 0 \)

### Step 8: Solve the equations
Subtract Equation 3 from Equation 1:
\[ (2x + 3y - z) - (2x - y + z) = 0 \implies 4y - 2z = 0 \implies 2y = z \quad \text{(Equation 4)} \]

From Equation 4, we can express \( z \) in terms of \( y \):
\[ z = 2y \]

### Step 9: Substitute \( z \) into Equation 1
Substituting \( z = 2y \) into Equation 1:
\[ 2x + 3y - 2y = 0 \implies 2x + y = 0 \implies x = -\frac{1}{2}y \quad \text{(Equation 5)} \]

### Step 10: Find ratios
Now we find the ratios:
1. \( \frac{x}{y} = -\frac{1}{2} \)
2. From Equation 4, \( \frac{y}{z} = \frac{1}{2} \) (since \( z = 2y \))
3. To find \( \frac{z}{x} \):
\[ z = 2y \implies \frac{z}{x} = \frac{2y}{-\frac{1}{2}y} = -4 \]

### Step 11: Calculate the expression
Now we need to calculate:
\[ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k \]
Substituting the values:
\[ -\frac{1}{2} + \frac{1}{2} - 4 + \frac{9}{2} \]
Calculating this:
\[ -\frac{1}{2} + \frac{1}{2} - 4 + \frac{9}{2} = 0 - 4 + \frac{9}{2} = -4 + 4.5 = \frac{1}{2} \]

### Final Answer
Thus, the value of \( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k \) is:
\[ \frac{1}{2} \]