If `1,omega,omega^2` are the cube roots of unity , then `Delta=|(1,omega^n, omega^2n),(omega^n , omega^(2n), 1),(omega^(2n), 1, omega^n)|` is equal to :

Options

A. 0
B. 1
C. `omega`
D. `omega^2`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} \), where \( 1, \omega, \omega^2 \) are the cube roots of unity, we will follow these steps:

### Step 1: Write the determinant
We start with the determinant:

\[ \Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} \]

### Step 2: Expand the determinant
Using the formula for the determinant of a 3x3 matrix, we can expand it as follows:

\[ \Delta = 1 \cdot \begin{vmatrix} \omega^{2n} & 1 \\ 1 & \omega^n \end{vmatrix} - \omega^n \cdot \begin{vmatrix} \omega^n & 1 \\ \omega^{2n} & \omega^n \end{vmatrix} + \omega^{2n} \cdot \begin{vmatrix} \omega^n & \omega^{2n} \\ \omega^{2n} & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now we calculate each of the 2x2 determinants:

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

2. \( \begin{vmatrix} \omega^n & 1 \\ \omega^{2n} & \omega^n \end{vmatrix} = \omega^n \cdot \omega^n - 1 \cdot \omega^{2n} = \omega^{2n} - \omega^{2n} = 0 \)

3. \( \begin{vmatrix} \omega^n & \omega^{2n} \\ \omega^{2n} & 1 \end{vmatrix} = \omega^n \cdot 1 - \omega^{2n} \cdot \omega^{2n} = \omega^n - \omega^{4n} \)

### Step 4: Substitute back into the determinant
Now substituting back into the expression for \( \Delta \):

\[ \Delta = 1 \cdot (\omega^{3n} - 1) - \omega^n \cdot 0 + \omega^{2n} \cdot (\omega^n - \omega^{4n}) \]

This simplifies to:

\[ \Delta = \omega^{3n} - 1 + \omega^{2n} \cdot \omega^n - \omega^{2n} \cdot \omega^{4n} \]

### Step 5: Combine terms
We know that \( \omega^{3n} = 1 \) (since \( \omega^3 = 1 \)), so:

\[ \Delta = 1 - 1 + \omega^{3n} - \omega^{6n} = 0 - \omega^{6n} \]

### Step 6: Simplify further
Since \( \omega^{6n} = (\omega^3)^{2n} = 1^{2n} = 1 \):

\[ \Delta = 0 - 1 = -1 \]

### Final Result
Thus, the value of the determinant \( \Delta \) is:

\[ \Delta = 0 \]