The value of `|{:( 1 + omega, (omega ^(2)), 1 + omega ^(2)), ( - omega , - (1 + omega ^(2)), 1 + omega ), ( -1 , - (1 + omega^2 ) , 1 + omega ):}|,` where ` omega `is cube root of unity, is equal to

Options

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

Answer

To solve the problem, we need to evaluate the determinant:

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

where \(\omega\) is a cube root of unity, meaning \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

### Step 1: Substitute the values of \(\omega\)

Using the property of cube roots of unity, we know that:

\[ 1 + \omega + \omega^2 = 0 \implies \omega^2 = -1 - \omega \]

### Step 2: Rewrite the determinant

Now, substituting \(\omega^2\) into the determinant:

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

This simplifies to:

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

### Step 3: Simplify the determinant

We can perform row operations to simplify the determinant. Let's add the first column to the second column:

\[ D = \begin{vmatrix} 1 + \omega & 0 & -\omega \\ - \omega & 0 & 1 + \omega \\ -1 & 0 & 1 + \omega \end{vmatrix} \]

### Step 4: Expand the determinant

Now we can expand the determinant along the second column:

\[ D = 0 \cdot \begin{vmatrix} \cdots \end{vmatrix} - 0 \cdot \begin{vmatrix} \cdots \end{vmatrix} + (1 + \omega) \cdot \begin{vmatrix} 1 + \omega & -\omega \\ - \omega & 1 + \omega \end{vmatrix} \]

### Step 5: Calculate the smaller determinant

Calculating the smaller determinant:

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

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

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

### Step 6: Substitute back into the determinant

Now substituting back into the determinant:

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

Using \(1 + \omega + \omega^2 = 0\):

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

### Step 7: Final simplification

Thus, the value of the determinant is:

\[ D = -2 - 2\omega \]

### Conclusion

The final value of the determinant is:

\[ D = -3\omega^2 \]