Solveeit Logo
Login

Question

Question: If $\omega$ is an imaginary cube root of unity, then the value of $\begin{bmatrix} 1 & \omega^2 & 1-...

If ω\omega is an imaginary cube root of unity, then the value of [1ω21−ω4ω11+ω51ωω2]\begin{bmatrix} 1 & \omega^2 & 1-\omega^4\\ \omega & 1 & 1+\omega^5\\ 1 & \omega & \omega^2 \end{bmatrix} is

A

-4

B

ω2−4\omega^2 - 4

C

ω2\omega^2

D

4

Answer

ω2−4\omega^2 - 4

Explanation

Solution

Let A=[1ω21−ω4ω11+ω51ωω2]A = \begin{bmatrix} 1 & \omega^2 & 1-\omega^4 \\ \omega & 1 & 1+\omega^5 \\ 1 & \omega & \omega^2 \end{bmatrix}, where ω\omega is an imaginary cube root of unity.

We use the properties of cube roots of unity: ω3=1\omega^3 = 1 and 1+ω+ω2=01 + \omega + \omega^2 = 0. Using ω3=1\omega^3 = 1, we have ω4=ω3⋅ω=1⋅ω=ω\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega and ω5=ω3⋅ω2=1⋅ω2=ω2\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2. Substitute these into the matrix: A=[1ω21−ωω11+ω21ωω2]A = \begin{bmatrix} 1 & \omega^2 & 1-\omega \\ \omega & 1 & 1+\omega^2 \\ 1 & \omega & \omega^2 \end{bmatrix}

Using 1+ω+ω2=01+\omega+\omega^2=0, we have 1+ω2=−ω1+\omega^2 = -\omega. So the matrix becomes: A=[1ω21−ωω1−ω1ωω2]A = \begin{bmatrix} 1 & \omega^2 & 1-\omega \\ \omega & 1 & -\omega \\ 1 & \omega & \omega^2 \end{bmatrix}

Now we calculate the determinant of this matrix. Let D=det⁡(A)D = \det(A). D=1⋅det⁡[1−ωωω2]−ω2⋅det⁡[ω−ω1ω2]+(1−ω)⋅det⁡[ω11ω]D = 1 \cdot \det \begin{bmatrix} 1 & -\omega \\ \omega & \omega^2 \end{bmatrix} - \omega^2 \cdot \det \begin{bmatrix} \omega & -\omega \\ 1 & \omega^2 \end{bmatrix} + (1-\omega) \cdot \det \begin{bmatrix} \omega & 1 \\ 1 & \omega \end{bmatrix}

Calculate the 2x2 determinants: det⁡[1−ωωω2]=(1)(ω2)−(−ω)(ω)=ω2+ω\det \begin{bmatrix} 1 & -\omega \\ \omega & \omega^2 \end{bmatrix} = (1)(\omega^2) - (-\omega)(\omega) = \omega^2 + \omega Since 1+ω+ω2=01 + \omega + \omega^2 = 0, we have ω2+ω=−1\omega^2 + \omega = -1.

det⁡[ω−ω1ω2]=(ω)(ω2)−(−ω)(1)=ω3+ω\det \begin{bmatrix} \omega & -\omega \\ 1 & \omega^2 \end{bmatrix} = (\omega)(\omega^2) - (-\omega)(1) = \omega^3 + \omega Since ω3=1\omega^3 = 1, this becomes 1+ω1 + \omega. Since 1+ω+ω2=01 + \omega + \omega^2 = 0, we have 1+ω=−ω21 + \omega = -\omega^2.

det⁡[ω11ω]=(ω)(ω)−(1)(1)=ω2−1\det \begin{bmatrix} \omega & 1 \\ 1 & \omega \end{bmatrix} = (\omega)(\omega) - (1)(1) = \omega^2 - 1.

Substitute these values back into the expression for DD: D=1⋅(−1)−ω2⋅(−ω2)+(1−ω)⋅(ω2−1)D = 1 \cdot (-1) - \omega^2 \cdot (-\omega^2) + (1-\omega) \cdot (\omega^2 - 1) D=−1+ω4+(1−ω)(ω2−1)D = -1 + \omega^4 + (1-\omega)(\omega^2 - 1) Using ω4=ω\omega^4 = \omega: D=−1+ω+(1−ω)(ω2−1)D = -1 + \omega + (1-\omega)(\omega^2 - 1)

Expand the last term: (1−ω)(ω2−1)=1⋅ω2−1⋅1−ω⋅ω2+ω⋅1(1-\omega)(\omega^2 - 1) = 1 \cdot \omega^2 - 1 \cdot 1 - \omega \cdot \omega^2 + \omega \cdot 1 =ω2−1−ω3+ω= \omega^2 - 1 - \omega^3 + \omega Using ω3=1\omega^3 = 1: =ω2−1−1+ω=ω2+ω−2= \omega^2 - 1 - 1 + \omega = \omega^2 + \omega - 2 Using ω2+ω=−1\omega^2 + \omega = -1: =−1−2=−3= -1 - 2 = -3.

Substitute this back into the expression for DD: D=−1+ω+(−3)D = -1 + \omega + (-3) D=ω−4D = \omega - 4

Applying 1+ω+ω2=01+\omega+\omega^2=0, ω=−1−ω2\omega = -1-\omega^2 D=−1−ω2−4=−5−ω2D = -1-\omega^2 - 4 = -5 - \omega^2 D=−5−ω2=−5−(−1−ω)=−5+1+ω=−4+ωD = -5 - \omega^2 = -5 - (-1-\omega) = -5 + 1 + \omega = -4 + \omega D=ω−4=−1−ω2−4=−5−ω2D = \omega - 4 = -1 - \omega^2 - 4 = -5 - \omega^2 D=−5−ω2D = -5 - \omega^2 ω2−4=−1−ω−4=−5−ω\omega^2 - 4 = -1-\omega - 4 = -5 - \omega Since 1+ω+ω2=01+\omega+\omega^2 = 0, ω2=−1−ω\omega^2 = -1-\omega Therefore, ω2−4=−1−ω−4=−5−ω\omega^2 - 4 = -1-\omega - 4 = -5-\omega

Thus, the value of the determinant is ω2−4\omega^2 - 4.