Understanding the Problem and Properties of \(\omega\)

The question asks us to evaluate a determinant involving complex numbers. We are given:

\( \omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \)

This value of \(\omega\) is a complex cube root of unity. It has important properties that we can use to simplify the calculation:

• Property 1: \(\omega^3 = 1\)
• Property 2: \(1 + \omega + \omega^2 = 0\)

From these fundamental properties, we can derive other useful relations:

• From \(1 + \omega + \omega^2 = 0\), we get \(\omega + \omega^2 = -1\).
• Also, \(1 + \omega = -\omega^2\).
• And \(1 + \omega^2 = -\omega\).
• Using \(\omega^3 = 1\), we can find the reciprocal values:
  • \(\frac{1}{\omega} = \frac{\omega^2}{\omega^3} = \omega^2\)
  • \(\frac{1}{\omega^2} = \frac{\omega}{\omega^3} = \omega\)

Simplifying the Determinant

The determinant we need to evaluate is:

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

Let's substitute the simplified properties into the matrix elements. Specifically, we substitute into the first and third rows:

• First row elements: \(1 + \omega = -\omega^2\), \(1 + \omega^2 = -\omega\), \(\omega + \omega^2 = -1\).
• Third row elements: \(\frac{1}{\omega} = \omega^2\), \(\frac{1}{\omega^2} = \omega\).

Substituting these into the determinant gives:

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

Calculating the Determinant Value

We can now evaluate this simplified determinant. A common technique is to use row or column operations to simplify the matrix before calculating the determinant.

Let's apply the column operation \(C_1 \rightarrow C_1 + C_2 + C_3\). This means we replace the first column with the sum of the elements in the first, second, and third columns.

The new elements in the first column (\(C_1'\)) will be:

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

Using the property \(1 + \omega + \omega^2 = 0\), all these new elements become 0:

• \(-(\omega^2 + \omega + 1) = -(0) = 0\)
• \(1 + \omega + \omega^2 = 0\)
• \(\omega^2 + \omega + 1 = 0\)

After applying the column operation, the determinant becomes:

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

A fundamental property of determinants is that if any column (or row) consists entirely of zeros, the determinant is equal to 0.

Therefore, \(D = 0\).

Conclusion

The value of the given determinant is 0.