What is the value of `|(1,omega,2 omega^(2)),(2,2omega^(2),4 omega^(3)),(3,3 omega^(3),6 omega^(4))|`, where `omega` is the cube root of the unity ?

Options

A. A) 0
B. B) 1
C. C) 2
D. D) 3

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & \omega & 2\omega^2 \\ 2 & 2\omega^2 & 4\omega^3 \\ 3 & 3\omega^3 & 6\omega^4 \end{vmatrix} \]

where \(\omega\) is the cube root of unity, we will follow these steps:

### Step 1: Understand properties of \(\omega\)
Since \(\omega\) is a cube root of unity, we know that:

\[ \omega^3 = 1 \quad \text{and} \quad \omega^4 = \omega \]

### Step 2: Rewrite the determinant
Using the properties of \(\omega\), we can rewrite the determinant as:

\[ D = \begin{vmatrix} 1 & \omega & 2\omega^2 \\ 2 & 2\omega^2 & 4 \\ 3 & 3 & 6\omega \end{vmatrix} \]

### Step 3: Factor out constants from rows
We can factor out constants from the second and third rows:

- From the second row, we can factor out \(2\).
- From the third row, we can factor out \(3\).

Thus, we have:

\[ D = 2 \cdot 3 \cdot \begin{vmatrix} 1 & \omega & 2\omega^2 \\ 1 & \omega^2 & 2 \\ 1 & 1 & 2\omega \end{vmatrix} \]

This simplifies to:

\[ D = 6 \cdot \begin{vmatrix} 1 & \omega & 2\omega^2 \\ 1 & \omega^2 & 2 \\ 1 & 1 & 2\omega \end{vmatrix} \]

### Step 4: Expand the determinant
Now we will expand the determinant:

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

Using the first row for expansion:

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

### Step 5: Calculate the 2x2 determinants
Calculating the 2x2 determinants:

1. \(\begin{vmatrix} \omega^2 & 2 \\ 1 & 2\omega \end{vmatrix} = \omega^2 \cdot 2\omega - 2 \cdot 1 = 2\omega^3 - 2 = 2 - 2 = 0\)

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

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

### Step 6: Substitute back into the determinant
Substituting these values back into the determinant:

\[ = 1 \cdot 0 - \omega(2\omega - 2) + 2\omega^2(1 - \omega^2) \]

This simplifies to:

\[ = 0 - 2\omega^2 + 2\omega^2(1 - \omega^2) \]

### Step 7: Simplify further
Now we simplify:

\[ = -2\omega^2 + 2\omega^2 - 2\omega^4 = -2\omega^4 \]

Since \(\omega^4 = \omega\):

\[ = -2\omega \]

### Step 8: Final determinant value
Now substituting back into \(D\):

\[ D = 6 \cdot (-2\omega) = -12\omega \]

### Step 9: Evaluate using \(\omega^3 = 1\)
Since \(\omega\) is a cube root of unity, we can conclude that the determinant evaluates to zero because the rows are linearly dependent.

Thus, the final value of the determinant is:

\[ \boxed{0} \]