If `omega` is a cube root of unity and `/_\ = |(1, 2 omega), (omega, omega^2)|`, then `/_\ ^2` is equal to (A) `- omega` (B) `omega` (C) `1` (D) `omega^2`

Options

A. `-omega`
B. `omega`
C. 1
D. `omega^2`

Answer

To solve the problem, we need to find the value of \(\Delta^2\) where \(\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}\) and \(\omega\) is a cube root of unity.

### Step 1: Calculate the Determinant \(\Delta\)

The determinant of a 2x2 matrix \(\begin{vmatrix} a & b \\ c & d \end{vmatrix}\) is calculated using the formula \(ad - bc\).

For our matrix:
\[ \Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix} \]

Calculating this gives:
\[ \Delta = (1)(\omega^2) - (2\omega)(\omega) = \omega^2 - 2\omega^2 = -\omega^2 \]

### Step 2: Calculate \(\Delta^2\)

Now, we need to find \(\Delta^2\):
\[ \Delta^2 = (-\omega^2)^2 \]

Calculating this gives:
\[ \Delta^2 = (-1)^2 \cdot (\omega^2)^2 = 1 \cdot \omega^4 \]

### Step 3: Simplify \(\omega^4\)

Since \(\omega\) is a cube root of unity, we know that \(\omega^3 = 1\). Therefore:
\[ \omega^4 = \omega^{3+1} = \omega^3 \cdot \omega = 1 \cdot \omega = \omega \]

### Final Result

Thus, we find that:
\[ \Delta^2 = \omega \]

The answer is (B) \(\omega\).

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