If `omega` is a complex cube root of unity, then the value of the determinant `|(1,omega,omega+1),(omega+1,1,omega),(omega,omega+1,1)|` is

Options

A. 0
B. `omega`
C. 2
D. 4

Answer

To solve the determinant \( D = \begin{vmatrix} 1 & \omega & \omega + 1 \\ \omega + 1 & 1 & \omega \\ \omega & \omega + 1 & 1 \end{vmatrix} \), where \( \omega \) is a complex cube root of unity, we will follow these steps:

### Step 1: Understanding the properties of \( \omega \)
The complex cube roots of unity satisfy the equation \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \). This means that \( \omega^2 = -1 - \omega \).

### Step 2: Simplifying the determinant
We can simplify the determinant by performing column operations. We will add all three columns together:

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:

\[ D = \begin{vmatrix} 1 + \omega + (\omega + 1) & \omega & \omega + 1 \\ (\omega + 1) + 1 + \omega & 1 & \omega \\ \omega + (\omega + 1) + 1 & \omega + 1 & 1 \end{vmatrix} \]

Calculating the first column:

- First row: \( 1 + \omega + \omega + 1 = 2\omega + 2 \)
- Second row: \( \omega + 1 + 1 + \omega = 2\omega + 2 \)
- Third row: \( \omega + \omega + 1 + 1 = 2\omega + 2 \)

Thus, the determinant becomes:

\[ D = \begin{vmatrix} 2\omega + 2 & \omega & \omega + 1 \\ 2\omega + 2 & 1 & \omega \\ 2\omega + 2 & \omega + 1 & 1 \end{vmatrix} \]

### Step 3: Factoring out common terms
Notice that the first column has a common factor of \( 2\omega + 2 \):

\[ D = (2\omega + 2) \begin{vmatrix} 1 & \omega & \omega + 1 \\ 1 & 1 & \omega \\ 1 & \omega + 1 & 1 \end{vmatrix} \]

### Step 4: Expanding the determinant
Now we need to compute the determinant:

\[ D' = \begin{vmatrix} 1 & \omega & \omega + 1 \\ 1 & 1 & \omega \\ 1 & \omega + 1 & 1 \end{vmatrix} \]

Using the first row to expand the determinant:

\[ D' = 1 \cdot \begin{vmatrix} 1 & \omega \\ \omega + 1 & 1 \end{vmatrix} - \omega \cdot \begin{vmatrix} 1 & \omega \\ 1 & 1 \end{vmatrix} + (\omega + 1) \cdot \begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} \]

Calculating these 2x2 determinants:

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

Putting this back into \( D' \):

\[ D' = 2 - \omega(1 - \omega) + (\omega + 1)(\omega - 1) \]

### Step 5: Simplifying \( D' \)
Now we simplify:

\[ D' = 2 - \omega + \omega^2 + \omega - 1 = 1 + \omega^2 \]

Using \( \omega^2 = -1 - \omega \):

\[ D' = 1 - 1 - \omega = -\omega \]

### Step 6: Final value of the determinant
Thus, we have:

\[ D = (2\omega + 2)(-\omega) = -2\omega^2 - 2\omega \]

Substituting \( \omega^2 = -1 - \omega \):

\[ D = -2(-1 - \omega) - 2\omega = 2 + 2\omega - 2\omega = 2 \]

### Conclusion
The value of the determinant is:

\[ \boxed{4} \]