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Mathematics Question on Determinants

Let Δ=∣111 1−1−w2w2 1ww4∣\Delta= \begin{vmatrix}1&1&1\\\ 1&-1-w^{2}&w^{2}\\\ 1&w&w^{4}\end{vmatrix}, where w≠1w \neq 1 is a complex number such that w3=1w^3 = 1. Then Δ\Delta equals

A

3w+w23 w + w^2

B

3w23w^2

C

3(w=w2)3(w = w^2)

D

−3w2-3 w^2

Answer

3w23w^2

Explanation

Solution

We have,
Δ=∣111 1−1−w2w2 1ww4∣\Delta=\begin{vmatrix}1 & 1 & 1 \\\ 1 & -1-w^{2} & w^{2} \\\ 1 & w & w^{4}\end{vmatrix}
=∣111 1ww2 1ww∣=\begin{vmatrix}1 & 1 & 1 \\\ 1 & w & w^{2} \\\ 1 & w & w\end{vmatrix}
[∵1+w+w2=0,w3=1]\left[\because 1+w+w^{2}=0, w^{3}=1\right]
=1(w2−w3)−1(w−w2)+1(w−w)=1\left(w^{2}-w^{3}\right)-1\left(w-w^{2}\right)+1(w-w)
=w2−1−w+w2=w^{2}-1-w+w^{2}
=2 w2−(1+w)=2 \,w^{2}-(1+w)
=2 w2−(−w2)=2\, w^{2}-\left(-w^{2}\right)
=3 w2=3 \,w^{2}