If -1 is a cube root of unity- then beginvmatrix 1 - 1- i - lomega - 2 - lomega - 2 11- i -1 - omega - 2-1 -i -i-lomega-1 -1 11 lendvmatrix equalsA- 0B- 1C- iD- -

The correct option is A 0Given det. = 1 i amp; ω+ω^2 amp;ω^2 1 1 i amp; 1 amp;ω^2 1 i amp; i+ω 1 amp; 1 (Using R_1 → R_1+R_3) (∵ R_1 an ...

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Byju's Answer

Standard XI

Mathematics

Sample Space

If ω=1 is a c...

Question

If $ω (\neq 1)$ is a cube root of unity, then
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + i + ω^{2} & ω^{2} 1 - i & - 1 & ω^{2} - 1 - i & - i + ω - 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ equals

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Solution

Given det. $= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 - i & ω + ω^{2} & ω^{2} - 1 1 - i & - 1 & ω^{2} - 1 - i & - i + ω - 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
(Using $R_{1} \to R_{1} + R_{3})$
$(∵ R_{1} a n d R_{2} a r e s a m e)$
$\Rightarrow$ value = 0

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If ω(≠1) is a cube root of unity, then ∣∣ ∣ ∣∣11+i+ω2ω21−i−1ω2−1−i−i+ω−1−1∣∣ ∣ ∣∣ equals

Given det. =∣∣ ∣ ∣∣1−iω+ω2ω2−11−i−1ω2−1−i−i+ω−1−1∣∣ ∣ ∣∣ (Using R1→R1+R3) (∵R1andR2 are same) ⇒ value = 0

If $ω (\neq 1)$ is a cube root of unity, then
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + i + ω^{2} & ω^{2} 1 - i & - 1 & ω^{2} - 1 - i & - i + ω - 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ equals

Given det. $= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 - i & ω + ω^{2} & ω^{2} - 1 1 - i & - 1 & ω^{2} - 1 - i & - i + ω - 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
(Using $R_{1} \to R_{1} + R_{3})$
$(∵ R_{1} a n d R_{2} a r e s a m e)$
$\Rightarrow$ value = 0