If $\displaystyle w$ is an imagenary cube root of unity , then find the value of

$\displaystyle

\begin{vmatrix}

1+w & w^2 & -w \\

1+w^2 & w & -w^2 \\

w^2+w & w & -w^2

\end{vmatrix}

$

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- May 3rd 2009, 10:11 PMzorroImagenary cube root Problems
If $\displaystyle w$ is an imagenary cube root of unity , then find the value of

$\displaystyle

\begin{vmatrix}

1+w & w^2 & -w \\

1+w^2 & w & -w^2 \\

w^2+w & w & -w^2

\end{vmatrix}

$ - May 4th 2009, 01:42 AMGamma
did you type it right? that bottom 2x2 minor has rows the same, is that correct?

If so, just do it, i recommend cofactor down the first column to take advantage of that being the same.

the first term is 0.

second is

$\displaystyle -(1+w^2)[-w+w^2]$

third is

$\displaystyle (w^2+w)[-w+w^2]$

Add them

$\displaystyle (-1-w^2 +w^2 +w)[-w+w^2]=(-1+w)[-w+w^2]=1-w^2-w^2+w=-2w^2+w+1$

But these are cube roots of unity, $\displaystyle w^2+w=-1 \Rightarrow w=-1-w^2$

$\displaystyle -2w^2+w+1= -2w^2 (-1 - w^2) + 1= -3w^3=-3$ - Dec 14th 2009, 01:16 PMzorroI didnt get the last part
- Dec 19th 2009, 07:52 PMzorroi have some doubt
- Dec 19th 2009, 08:29 PMzorroCould u please provide me with links