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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

ω²

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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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Similar questions

ω

is a cube root of unity and

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

Δ^{2}

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 an imaginary cube root of unity, then the value

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + ω & ω^{2} & - ω 1 + ω^{2} & ω & - ω^{2} ω^{2} + ω & ω & - ω^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

ω

is imaginary cube root of unity then prove that the value of

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + ω & ω^{2} & - ω 1 + ω^{2} & ω & - ω^{2} ω^{2} + ω & ω & - ω^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

- 3 ω^{2}

.

ω \neq 1

is a cube root of unity and

△ = ∣ ∣ ∣ ∣ ∣ \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

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