The value of `Delta=|(1,1,1),(1,omega,omega^2),(1,omega^2,omega)|` is (where `omega=(-1+sqrt3i)/2`)

Options

A. `3sqrt3i`
B. `-3sqrt3i`
C. `-sqrt3i`
D. `sqrt3i`

Answer

To find the value of the determinant \( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix} \), where \( \omega = \frac{-1 + \sqrt{3}i}{2} \), we will follow these steps:

### Step 1: Calculate \( \omega^2 \)
First, we need to calculate \( \omega^2 \):
\[ \omega = \frac{-1 + \sqrt{3}i}{2} \]
Using the formula for squaring a complex number:
\[ \omega^2 = \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 = \frac{(-1)^2 + 2(-1)(\sqrt{3}i) + (\sqrt{3}i)^2}{4} = \frac{1 - 2\sqrt{3}i - 3}{4} = \frac{-2 - 2\sqrt{3}i}{4} = \frac{-1 - \sqrt{3}i}{2} \]

### Step 2: Set up the determinant
Now we can set up the determinant:
\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix} \]

### Step 3: Expand the determinant
Using the determinant formula for a 3x3 matrix:
\[ \Delta = 1 \cdot \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & \omega \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & \omega^2 \\ 1 & \omega \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & \omega \\ 1 & \omega^2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & \omega \end{vmatrix} = \omega \cdot \omega - \omega^2 \cdot \omega^2 = \omega^2 - \omega^4 \)
2. \( \begin{vmatrix} 1 & \omega^2 \\ 1 & \omega \end{vmatrix} = 1 \cdot \omega - 1 \cdot \omega^2 = \omega - \omega^2 \)
3. \( \begin{vmatrix} 1 & \omega \\ 1 & \omega^2 \end{vmatrix} = 1 \cdot \omega^2 - 1 \cdot \omega = \omega^2 - \omega \)

Thus,
\[ \Delta = \omega^2 - \omega^4 - (\omega - \omega^2) + (\omega^2 - \omega) \]
Simplifying this expression:
\[ \Delta = \omega^2 - \omega^4 - \omega + \omega^2 + \omega^2 - \omega = 3\omega^2 - \omega - \omega^4 \]

### Step 4: Substitute \( \omega^4 \)
Since \( \omega^3 = 1 \) (as \( \omega \) is a cube root of unity), we have \( \omega^4 = \omega \). Therefore:
\[ \Delta = 3\omega^2 - \omega - \omega = 3\omega^2 - 2\omega \]

### Step 5: Substitute \( \omega \) and \( \omega^2 \)
Now substituting \( \omega \) and \( \omega^2 \):
\[ \Delta = 3\left(\frac{-1 - \sqrt{3}i}{2}\right) - 2\left(\frac{-1 + \sqrt{3}i}{2}\right) \]
Calculating this:
\[ = \frac{-3 - 3\sqrt{3}i + 2 - 2\sqrt{3}i}{2} = \frac{-1 - 5\sqrt{3}i}{2} \]

### Final Answer
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = \frac{-1 - 5\sqrt{3}i}{2} \]