Let ω=-(1/2)+(√3/2) i. Then the value of the determinant Δ=|1 1 1 1 -1-ω2 ω2 1 ω2 ω4| is

Question

Let ω=-(1/2)+(√3/2) i. Then the value of the determinant Δ=|1 1 1 1 -1-ω2 ω2 1 ω2 ω4| is (A) 3 ω (B) 3 ω(ω-1) (C) 3 ω2 (D) 3 ω(1-ω)

Tardigrade Admin · Accepted Answer

Using 1+ω2=-ω, ω4=ω and applying C2 arrow C2-C1, C3 arrow C3-C1 we get Δ =|1 0 0 1 ω-1 ω2-1 1 ω2-1 ω-1| =(ω-1)2-(ω2-1)2 =(ω+ω2-2)(ω-1-ω2+1) =(-3)(ω-ω2)=3 ω(ω-1)

Q. Let $ω = - \frac{1}{2} + \frac{3}{2} i$ . Then the value of the determinant $Δ = ∣ ∣ 111 1 - 1 - ω^{2} ω^{2} 1 ω^{2} ω^{4} ∣ ∣$ is

$3 ω$

$3 ω (ω - 1)$

$3 ω^{2}$

$3 ω (1 - ω)$

Using $1 + ω^{2} = - ω, ω^{4} = ω$ and applying $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$ we get
$Δ = ∣ ∣ 111 0 ω - 1 ω^{2} - 1 0 ω^{2} - 1 ω - 1 ∣ ∣$
$= (ω - 1)^{2} - (ω^{2} - 1)^{2}$
$= (ω + ω^{2} - 2) (ω - 1 - ω^{2} + 1)$
$= (- 3) (ω - ω^{2}) = 3 ω (ω - 1)$

Q. Let ω=−21​+23​​i. Then the value of the determinant Δ=∣∣​111​1−1−ω2ω2​1ω2ω4​∣∣​ is

3ω

3ω(ω−1)

3ω2

3ω(1−ω)