Let ω=-(1/2)+i(√ 3/2), then value of the determinant | 1 1& 1 1 -1-ω2 ω2 1 ω2 ω | is

Question

Let ω=-(1/2)+i(√ 3/2), then value of the determinant | 1 1& 1 1 -1-ω2 ω2 1 ω2 ω | is  (A) 3 ω (B) 3 ω(ω-1) (C) 3 ω2 (D) 3 ω(1- ω)

Tardigrade Admin · Accepted Answer

Let Δ=|1 1& 1 1 -1-ω2 ω2 1 ω2 ω | is Applying R2arrow R2-R1;R3arrow R3-R1 =|1 1& 1 0 -2-ω2 ω2-1 0 ω2-1 ω-1 | =(-2-ω2)(ω-1)-(ω2-1)2 =-2ω+2-ω3+ω2-(ω4-2ω2+1) =3ω2-3ω=3ω(ω-1) [∵ ω4=ω]

Q. Let $ω = - \frac{1}{2} + i \frac{3}{2},$ then value of the determinant $∣ ∣ 111 1 - 1 - ω^{2} ω^{2} 1 ω^{2} ω ∣ ∣ i s$

$3 ω$

$3 ω (ω - 1)$

$3 ω^{2}$

$3 ω (1 - ω)$

Q. Let ω=−21​+i23​​, then value of the determinant ∣∣​111​1−1−ω2ω2​1ω2ω​∣∣​is

3ω

3ω(ω−1)

3ω2

3ω(1−ω)

Let Δ=∣∣​111​1−1−ω2ω2​1ω2ω​∣∣​is Applying R2​→R2​−R1​;R3​→R3​−R1​ =∣∣​100​1−2−ω2ω2−1​1ω2−1ω−1​∣∣​ =(−2−ω2)(ω−1)−(ω2−1)2 =−2ω+2−ω3+ω2−(ω4−2ω2+1) =3ω2−3ω=3ω(ω−1)[∵ω4=ω]

Q. Let $ω = - \frac{1}{2} + i \frac{3}{2},$ then value of the determinant $∣ ∣ 111 1 - 1 - ω^{2} ω^{2} 1 ω^{2} ω ∣ ∣ i s$

$3 ω$

$3 ω (ω - 1)$

$3 ω^{2}$

$3 ω (1 - ω)$