If x,y,z are complex numbers , then `Delta=|(0,-y,-z),(bary,0,-x),(barz,barx,0)|` is equal to

Options

A. purely real
B. purely imaginary
C. complex
D. 0

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ \Delta = \begin{vmatrix} 0 & -y & -z \\ \bar{y} & 0 & -x \\ \bar{z} & \bar{x} & 0 \end{vmatrix} \]

### Step 1: Expand the Determinant

We can use the cofactor expansion along the first row:

\[ \Delta = 0 \cdot \begin{vmatrix} 0 & -x \\ \bar{x} & 0 \end{vmatrix} - (-y) \cdot \begin{vmatrix} \bar{y} & -x \\ \bar{z} & 0 \end{vmatrix} - (-z) \cdot \begin{vmatrix} \bar{y} & 0 \\ \bar{z} & \bar{x} \end{vmatrix} \]

This simplifies to:

\[ \Delta = y \cdot \begin{vmatrix} \bar{y} & -x \\ \bar{z} & 0 \end{vmatrix} + z \cdot \begin{vmatrix} \bar{y} & 0 \\ \bar{z} & \bar{x} \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

Now, we need to calculate the two 2x2 determinants.

1. For the first determinant:

\[ \begin{vmatrix} \bar{y} & -x \\ \bar{z} & 0 \end{vmatrix} = \bar{y} \cdot 0 - (-x) \cdot \bar{z} = x \bar{z} \]

2. For the second determinant:

\[ \begin{vmatrix} \bar{y} & 0 \\ \bar{z} & \bar{x} \end{vmatrix} = \bar{y} \cdot \bar{x} - 0 \cdot \bar{z} = \bar{y} \bar{x} \]

### Step 3: Substitute Back into the Expression for Delta

Now substitute these results back into the expression for \(\Delta\):

\[ \Delta = y \cdot (x \bar{z}) + z \cdot (\bar{y} \bar{x}) \]

This simplifies to:

\[ \Delta = yx \bar{z} + z \bar{y} \bar{x} \]

### Final Result

Thus, the value of \(\Delta\) is:

\[ \Delta = yx \bar{z} + z \bar{y} \bar{x} \]