If \(\rm \omega=-\frac{1}{2}+i\frac{\sqrt3}{2}\) then what is \(\rm \begin{vmatrix}1+\omega &1+\omega^2&\omega+\omega^2\\\ 1&\omega\ & \omega^2\\\ \frac{1}{\omega}&\frac{1}{\omega^2}&1\end{vmatrix}\) equal to? 

Concept:

Determinant and Column Operations:

• The determinant of a matrix is a scalar value representing area, volume, or invertibility.
• Column operations like \( C_1 \rightarrow C_1 + C_2 + C_3 \) can simplify a determinant.
• If one column (or row) becomes a linear combination of others or zero, the determinant becomes 0.
• \( \omega \) is a cube root of unity with identity: \( 1 + \omega + \omega^2 = 0 \).

Calculation:

Let \( I = \begin{bmatrix} 1+\omega & 1+\omega^2 & \omega + \omega^2 \\ 1 & \omega & \omega^2 \\ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \end{bmatrix} \)

⇒ Apply column operation: \( C_1 \rightarrow C_1 + C_2 + C_3 \)

⇒ First column becomes:

\( C_1 = \begin{bmatrix} (1+\omega) + (1+\omega^2) + (\omega + \omega^2) \\ 1 + \omega + \omega^2 \\ \frac{1}{\omega} + \frac{1}{\omega^2} + 1 \end{bmatrix} \)

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

⇒ First column becomes: \( \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \)

⇒ New matrix:

\( I = \begin{bmatrix} 0 & 1+\omega^2 & \omega + \omega^2 \\ 0 & \omega & \omega^2 \\ 0 & \frac{1}{\omega^2} & 1 \end{bmatrix} \)

⇒ Since first column is all zero,

∴ Determinant I = 0