If ω≠$$1is$$ a complex cube root of unity, and $$x+iy=1i$$−ω−$$i1$$ω$$2$$ω−ω$$21$$ then

Correct option is 1x = – 1, y = 0

Questions

If $ω \neq 1$ is a complex cube root of unity, and $x + i y = |\begin{array}{ccc} 1 & i & - ω \\ - i & 1 & ω^{2} \\ ω & - ω^{2} & 1 \end{array}|$ then

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x = – 1, y = 0

x = 1, y = – 1

x = 1, y = 1

none of these

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detailed solution

Correct option is A

Applying C1→C1+iC2, we getx+iy=0i−ω01ω2ω−iω2−ω21=ω−iω2iω2+ω=ω2−i2ω4=ω2+ω=−1⇒x=−1,y=0.

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If ω≠1is a complex cube root of unity, and x+iy=1i−ω−i1ω2ω−ω21 then

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If $ω \neq 1$ is a complex cube root of unity, and $x + i y = |\begin{array}{ccc} 1 & i & - ω \\ - i & 1 & ω^{2} \\ ω & - ω^{2} & 1 \end{array}|$ then