Multiple choice

Let Δ = $\begin{vmatrix} \ a & a+b & a+b+c \ \ 3a & 4a+3b & 5a+4b+3c \ \ 6a & 9a + 6b & 11a + 9b + 6c \ \end{vmatrix}$, where a = i, b = $\omega$, c = $\omega$2, then Δ is equal to

  1. -i
    • 1

  3. 1

  4. i
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A Correct answer
Explanation

Solving the determinant by reducing the rows and columns:

Expanding in column 1, we get determinant = a3 = i3 = -i [i x i = -1]

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