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

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

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

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 $ω \neq 1$ 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

answer is A.

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Detailed Solution

Applying $C_{1} \to C_{1} + i C_{2}$ , we get

$\begin{array}{l} x + i y = |\begin{array}{ccc} 0 & i & - ω \\ 0 & 1 & ω^{2} \\ ω - i ω^{2} & - ω^{2} & 1 \end{array}| \\ = (ω - i ω^{2}) (i ω^{2} + ω) \\ = ω^{2} - i^{2} ω^{4} = ω^{2} + ω = - 1 \\ \Rightarrow x = - 1, y = 0 . \end{array}$