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Mathematics Question on Algebra of Complex Numbers

Let ω=−12+i32,\omega=-\frac{1}{2}+i\frac{\sqrt 3}{2}, then value of the determinant ∣111 1−1−ω2ω2 1ω2ω ∣is\begin {vmatrix} 1 & 1& 1 \\\ 1 & -1-\omega^2 & \omega^2 \\\ 1 & \omega^2 & \omega \\\ \end {vmatrix} is

A

3ω3 \omega

B

3ω(ω−1)3 \omega(\omega-1)

C

3ω23 \omega^2

D

3ω(1−ω)3 \omega(1- \omega)

Answer

3ω(ω−1)3 \omega(\omega-1)

Explanation

Solution

Let δ=\delta= ∣111 1−1−ω2ω2 1ω2ω ∣is\begin {vmatrix} 1 & 1& 1 \\\ 1 & -1-\omega^2 & \omega^2 \\\ 1 & \omega^2 & \omega \\\ \end {vmatrix} is
Applying R2→R@−R1;R3→R3−R1R_2\rightarrow R_@-R_1;R_3\rightarrow R_3-R_1
=∣111 0−2−ω2ω2−1 0ω2−1ω−1 ∣\begin {vmatrix} 1 & 1& 1 \\\ 0 & -2-\omega^2 & \omega^2-1 \\\ 0 & \omega^2-1 & \omega-1 \\\ \end {vmatrix}
=(−2−ω2)(ω−1)−(ω2−1)2=(-2-\omega^2)(\omega-1)-(\omega^2-1)^2
=−2ω+2−ω3+ω2−(ω4−2ω2+1)=-2\omega+2-\omega^3+\omega^2-(\omega^4-2\omega^2+1)
=3ω2−3ω=3ω(ω−1)          [∵ω4=ω]=3\omega^2-3\omega=3\omega(\omega-1) \ \ \ \ \ \ \ \ \ \ [\because \omega^4=\omega]