The value `|(1 + omega,omega^(2),-omega),(1 + omega^(2),omega,-omega^(2)),(omega^(2)+omega,omega,-omega^(2))|` is equal to ( ` omega` is an imaginary cube root of unity)

Options

A. 0
B. ` 2 omega`
C. ` 2 omega^(2)`
D. `- 3 omega^(2)`

Answer

To find the value of the determinant

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

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

### Step 1: Recall properties of \(\omega\)
We know that \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\). From this, we can express \(\omega^2\) as \(-1 - \omega\).

### Step 2: Expand the determinant
We will expand the determinant along the first row:

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

### Step 3: Calculate the 2x2 determinants
1. The first 2x2 determinant is:

\[ \begin{vmatrix} \omega & -\omega^2 \\ \omega & -\omega^2 \end{vmatrix} = \omega(-\omega^2) - (-\omega^2)(\omega) = -\omega^3 + \omega^3 = 0 \]

2. The second 2x2 determinant is:

\[ \begin{vmatrix} 1 + \omega^2 & -\omega^2 \\ \omega^2 + \omega & -\omega^2 \end{vmatrix} = (1 + \omega^2)(-\omega^2) - (-\omega^2)(\omega^2 + \omega) \]
\[ = -\omega^2 - \omega^4 + \omega^4 + \omega^3 = -\omega^2 + 1 \]

3. The third 2x2 determinant is:

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

### Step 4: Substitute back into the determinant
Now substituting back, we have:

\[ D = (1 + \omega)(0) - \omega^2(-\omega^2 + 1) + \omega(\omega - \omega^2) \]
\[ = 0 + \omega^2(\omega^2 - 1) + \omega^2 - \omega^3 \]
\[ = \omega^4 - \omega^2 + \omega^2 - 1 = \omega^4 - 1 \]

### Step 5: Simplify using \(\omega^3 = 1\)
Since \(\omega^4 = \omega\), we have:

\[ D = \omega - 1 \]

### Step 6: Final result
Thus, the value of the determinant is:

\[ D = -\omega^2 \]