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Mathematics Question on Determinants

If ω\omega is an imaginary cube root of unity, then the value of the determinant ∣1+ωω2−ω 1+ω2ω−ω2 ω+ω2ω−ω2∣\begin{vmatrix}1+\omega&\omega^{2}&-\omega\\\ 1+\omega^{2}&\omega&-\omega^{2}\\\ \omega+\omega^2&\omega&-\omega^{2}\end{vmatrix}

A

−2ω-2\omega

B

−3ω2-3\omega^2

C

−1-1

D

00

Answer

−3ω2-3\omega^2

Explanation

Solution

∣1+ωω2−ω 1+ω2ω−ω2 ω+ω2ωω2∣\begin{vmatrix}1+\omega&\omega^{2}&-\omega\\\ 1+\omega^{2}&\omega&-\omega^{2}\\\ \omega+\omega^{2}&\omega& \omega^{2}\end{vmatrix}
=∣0ω2−ω 0ω−ω2 −1+ωω−ω2∣=\begin{vmatrix}0&\omega^{2}&-\omega\\\ 0&\omega&-\omega^{2}\\\ -1+\omega&\omega&-\omega^{2}\end{vmatrix}
[∵C1→C1+C2]\left[\because C_{1} \rightarrow C_{1}+C_{2}\right]
=(−1+ω)(−ω4+ω2)=(ω−1)(ω2−ω)=(-1+\omega)\left(-\omega^{4}+\omega^{2}\right)=(\omega-1)\left(\omega^{2}-\omega\right)
=ω3−ω2−ω2+ω=−3ω2=\omega^{3}-\omega^{2}-\omega^{2}+\omega=-3 \omega^{2}