Question

Let \( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1-w^2 & w^2 \\ 1 & w & w^4 \end{vmatrix} \), where \( w \neq 1 \) is a complex number such that \( w^3 = 1 \). Then \( \Delta \) equals:

• A. \( 3w + w^2 \)
• B. \( 3w^2 \)
• C. \( 3(w - w^2) \)
• D. \( -3w^2 \)
• E. \( 3w^2 + 1 \)

Hint:

For cube roots of unity, reduce all higher powers using \(w^3=1\) and simplify expressions using \(1+w+w^2=0\).

Answer 1

The Correct Option is B

Concept: For cube roots of unity, \[ w^3 = 1, \qquad 1+w+w^2 = 0 \] Using these identities: \[ -1-w^2 = w, \qquad w^4 = w \]

Step 1: Simplify the determinant.
Substituting the identities: \[ \Delta= \begin{vmatrix} 1 & 1 & 1 \\ 1 & w & w^2 \\ 1 & w & w \end{vmatrix} \] Apply row operations: \[ R_2 \to R_2-R_1, \qquad R_3 \to R_3-R_1 \] \[ \Delta= \begin{vmatrix} 1 & 1 & 1 \\ 0 & w-1 & w^2-1 \\ 0 & w-1 & w-1 \end{vmatrix} \]

Step 2: Expand the determinant.
Expanding along the first column: \[ \Delta= \begin{vmatrix} w-1 & w^2-1 \\ w-1 & w-1 \end{vmatrix} \] \[ = (w-1)^2-(w-1)(w^2-1) \] Factor out \((w-1)\): \[ = (w-1)\big[(w-1)-(w^2-1)\big] \] \[ = (w-1)(w-w^2) \] \[ = w(1-w)(w-1) \] \[ = -w(w-1)^2 \] Now, \[ (w-1)^2=w^2-2w+1 \] Using \[ 1+w+w^2=0 \Rightarrow 1+w^2=-w \] \[ (w-1)^2=-3w \] Therefore, \[ \Delta=-w(-3w)=3w^2 \] \[ \boxed{3w^2} \]