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. 
 

Answer 2

This determinant can be evaluated using the row-based property that a determinant is zero whenever two rows are identical (or proportional). Let's verify this by checking each option against the actual structure of the matrix.

1. 1: A determinant of a matrix with repeated rows can never be a nonzero value like 1; any row operation (subtracting one identical row from another) immediately produces a zero row, which forces the determinant itself to be zero.
2. -1: Same reasoning as above rules this out - a nonzero determinant is impossible once two (in fact all three) rows are identical.
3. 0: Looking at the matrix, all three rows are exactly \( (\omega^2, \omega, \omega^2) \) - they are identical. Subtracting row 1 from row 2 and from row 3 turns both into zero rows, and a determinant with a zero row is always 0, regardless of what \( \omega \) is.
4. \( \omega \): This would require the determinant to depend on \( \omega \) in a nontrivial way, but since the rows are identical the determinant collapses to zero identically, independent of the specific value of \( \omega \).

Since all three rows of the given determinant are exactly the same, the determinant is identically zero, with no dependence on \( \omega \) at all.

Therefore, the correct answer is 0.