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Question: If $\omega$ is cube root of unity, then value of the determinant $\begin{vmatrix} 1 & 1 & 1 \\ 1 & \...

If ω\omega is cube root of unity, then value of the determinant ∣1111ω2ω1ωω2∣=\begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{vmatrix} =

A

-3

B

33i3\sqrt{3}i

C

3

D

−33i-3\sqrt{3}i

Answer

33i3\sqrt{3}i

Explanation

Solution

The given determinant is: D=∣1111ω2ω1ωω2∣D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{vmatrix} where ω\omega is a cube root of unity.

We know the following properties of cube roots of unity:

  1. ω3=1\omega^3 = 1
  2. 1+ω+ω2=01 + \omega + \omega^2 = 0

We can evaluate the determinant using cofactor expansion along the first row: D=1⋅∣ω2ωωω2∣−1⋅∣1ω1ω2∣+1⋅∣1ω21ω∣D = 1 \cdot \begin{vmatrix} \omega^2 & \omega \\ \omega & \omega^2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & \omega \\ 1 & \omega^2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & \omega^2 \\ 1 & \omega \end{vmatrix} D=(ω2⋅ω2−ω⋅ω)−(1⋅ω2−1⋅ω)+(1⋅ω−1⋅ω2)D = (\omega^2 \cdot \omega^2 - \omega \cdot \omega) - (1 \cdot \omega^2 - 1 \cdot \omega) + (1 \cdot \omega - 1 \cdot \omega^2) D=(ω4−ω2)−(ω2−ω)+(ω−ω2)D = (\omega^4 - \omega^2) - (\omega^2 - \omega) + (\omega - \omega^2) Since ω3=1\omega^3 = 1, we have ω4=ω3⋅ω=1⋅ω=ω\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega. Substitute ω4=ω\omega^4 = \omega into the expression: D=(ω−ω2)−(ω2−ω)+(ω−ω2)D = (\omega - \omega^2) - (\omega^2 - \omega) + (\omega - \omega^2) Now, combine the terms: D=ω−ω2−ω2+ω+ω−ω2D = \omega - \omega^2 - \omega^2 + \omega + \omega - \omega^2 D=3ω−3ω2D = 3\omega - 3\omega^2 D=3(ω−ω2)D = 3(\omega - \omega^2) To find the value of ω−ω2\omega - \omega^2, we use the standard values of ω\omega: ω=ei2π/3=cos⁡(2π/3)+isin⁡(2π/3)=−12+i32\omega = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2} ω2=ei4π/3=cos⁡(4π/3)+isin⁡(4π/3)=−12−i32\omega^2 = e^{i4\pi/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}

Now, calculate ω−ω2\omega - \omega^2: ω−ω2=(−12+i32)−(−12−i32)\omega - \omega^2 = \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) ω−ω2=−12+i32+12+i32\omega - \omega^2 = -\frac{1}{2} + i\frac{\sqrt{3}}{2} + \frac{1}{2} + i\frac{\sqrt{3}}{2} ω−ω2=2i32=i3\omega - \omega^2 = 2i\frac{\sqrt{3}}{2} = i\sqrt{3} Substitute this value back into the expression for DD: D=3(i3)=33iD = 3(i\sqrt{3}) = 3\sqrt{3}i

Therefore, the value of the determinant is 33i3\sqrt{3}i.