Question

The value of the determinant where \( \omega \) is cube root of unity is \[ \begin{vmatrix} \omega^2 & \omega & \omega^2 \\ \omega^2 & \omega & \omega^2 \\ \omega^2 & \omega & \omega^2 \end{vmatrix} \]

• A. 1
• B. -1
• C. 0
• D. \( \omega \)

Hint:

If any two rows or columns of a matrix are identical, the determinant will be zero.

Answer 1

The Correct Option is C

Step 1: Properties of cube roots of unity. 
The cube roots of unity satisfy the equation \( \omega^3 = 1 \), and the properties are: \[ 1 + \omega + \omega^2 = 0 \quad \text{and} \quad \omega^2 = \omega^{-1} \] Step 2: Simplifying the determinant. 
The given matrix is of the form: \[ \begin{vmatrix} \omega^2 & \omega & \omega^2 \\ \omega^2 & \omega & \omega^2 \\ \omega^2 & \omega & \omega^2 \end{vmatrix} \] Since the rows are identical, the determinant will be 0. Thus, the value of the determinant is \( 0 \). 
Step 3: Conclusion. 
The correct answer is (3) 0.