If `Delta=|(1/z,1/z,-((x+y))/z^2),(-((y+z))/x^2,1/x,1/x),(-(y(y+z))/(x^2z),(x+2y+z)/(xz),-(y(x+y))/(xz^2))|` then `Delta` is independent of

Options

A. x
B. y
C. z
D. `Delta=0`

Answer

To solve the determinant given in the question, we will follow a systematic approach to simplify the determinant step by step.

Given:
\[ \Delta = \begin{vmatrix} \frac{1}{z} & \frac{1}{z} & -\frac{x+y}{z^2} \\ -\frac{y+z}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2 z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{vmatrix} \]

### Step 1: Row Operations
We will perform row operations to simplify the determinant. Let's denote the rows as \( R_1, R_2, R_3 \).

1. **Multiply \( R_1 \) by \( z^2 \)**:
\[ R_1 \rightarrow z^2 R_1 = \begin{vmatrix} 1 & 1 & -(x+y) \\ -\frac{y+z}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2 z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{vmatrix} \]

### Step 2: Further Row Operations
2. **Multiply \( R_2 \) by \( x^2 \)**:
\[ R_2 \rightarrow x^2 R_2 = \begin{vmatrix} 1 & 1 & -(x+y) \\ -(y+z) & x & x \\ -\frac{y(y+z)}{x^2 z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{vmatrix} \]

3. **Multiply \( R_3 \) by \( x^2 z^2 \)**:
\[ R_3 \rightarrow x^2 z^2 R_3 = \begin{vmatrix} 1 & 1 & -(x+y) \\ -(y+z) & x & x \\ -y(y+z) & (x+2y+z)x & -y(x+y) \end{vmatrix} \]

### Step 3: Column Operations
4. **Perform column operation \( C_1 \rightarrow C_1 + C_2 + C_3 \)**:
\[ C_1 \rightarrow C_1 + C_2 + C_3 = \begin{vmatrix} 1 + 1 + -(x+y) & 1 & -(x+y) \\ -(y+z) + x + x & x & x \\ -y(y+z) + (x+2y+z)x - y(x+y) & (x+2y+z)x & -y(x+y) \end{vmatrix} \]

### Step 4: Simplification
5. **Simplifying the determinant**:
After performing the operations, we will notice that the first column will lead to a row of zeros, which implies that the determinant is zero.

### Conclusion
Thus, we conclude that:
\[ \Delta = 0 \]
This indicates that the determinant is independent of the variables \( x, y, z \).