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Evaluation of a Determinant

If ω is a cub...

Question

If $ω$ is a cube root of unity and $Δ = ∣ ∣ ∣ \begin{matrix} 1 & 2 ω ω & ω^{2} \end{matrix} ∣ ∣ ∣$ , then $Δ^{2}$ is equal to

A

- ω

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B

ω^{2}

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C

1

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D

ω

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Solution

The correct option is D $ω$
Since $Δ = ω^{2} - 2 ω^{2} = - ω^{2} .$ Therefore $Δ^{2} = ω^{4} = ω .$

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ω

is a cube root of unity and

Δ = ∣ ∣ ∣ \begin{matrix} 1 & 2 ω ω & ω^{2} \end{matrix} ∣ ∣ ∣

Δ^{2}

1, ω

ω^{2}

are the cube roots of unity, then

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & ω^{n} & ω^{2 n} ω^{n} & ω^{2 n} & 1 ω^{2 n} & 1 & ω^{n} \end{matrix} ∣ ∣ ∣ ∣ ∣

ω

is cube root of unity, then

Δ = ∣ ∣ ∣ ∣ \begin{matrix} x + 1 ω ω^{2} \end{matrix} \begin{matrix} ω x + ω^{2} 1 \end{matrix} \begin{matrix} ω^{2} 1 x + ω \end{matrix} ∣ ∣ ∣ ∣ =

ω \neq 1

is cube root of unity and

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x + ω^{2} & ω & 1 ω & ω^{2} & 1 + x 1 & x + ω & ω^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

, then the value of

x

ω

is a non-real cube root of unity and n is not a multiple of 3, then

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & ω^{n} & ω^{2 n} ω^{2 n} & 1 & ω^{n} ω^{n} & ω^{2 n} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

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