What is the value of `|{:(" "1-i," "omega^(2)," "-omega),(" "omega^(2)+i," "omega," "-i),(1-2i-omega^(2),omega^(2)-omega,i-omega):}|`, where `omega` is the cube root of unity ?

Options

A. -1
B. 1
C. 2
D. 0

Explanation

To find the value of the determinant

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

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

### Step 1: Understand the properties of \(\omega\)
The cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \]
and
\[ \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \]
with the property that \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

### Step 2: Simplify the determinant
We can perform row operations to simplify the determinant. Let's subtract the second row from the first row and then subtract the third row from the result:

\[ R_1 = R_1 - R_2 \]

This gives us:

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

Calculating the new elements of the first row:

1. First element: \( (1 - i) - (\omega^2 + i) = 1 - \omega^2 - 2i \)
2. Second element: \( \omega^2 - \omega \)
3. Third element: \( -\omega + i \)

So we have:

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

### Step 3: Further simplify the determinant
Next, we can perform another operation to simplify further. Let's subtract the first row from the third row:

\[ R_3 = R_3 - R_1 \]

This gives us:

1. First element: \( (1 - 2i - \omega^2) - (1 - \omega^2 - 2i) = 0 \)
2. Second element: \( (\omega^2 - \omega) - (\omega^2 - \omega) = 0 \)
3. Third element: \( (i - \omega) - (-\omega + i) = 0 \)

Thus, the third row becomes:

\[ \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \]

### Step 4: Conclusion
Since one row of the determinant is all zeros, the value of the determinant is:

\[ D = 0 \]

### Final Answer
The value of the determinant is \(0\).