If w is a complex cube root of unity, show that

Question:

If w is a complex cube root of unity, show that

$\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

Solution:

Here,

$\mathrm{LHS}=\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]$

$=\left[\begin{array}{ccc}1+w & w+w^{2} & w^{2}+1 \\ w+w^{2} & w^{2}+1 & 1+w \\ w^{2}+w & 1+w^{2} & w+1\end{array}\right]\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]$

$=\left[\begin{array}{ccc}-w^{2} & -1 & -w \\ -1 & -w & -w^{2} \\ -1 & -w & -w^{2}\end{array}\right]\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]$      $\left(\because 1+w+w^{2}=0\right.$ and $\left.w^{3}=1\right)$

$=\left[\begin{array}{c}-w^{2}-w-w^{3} \\ -1-w^{2}-w^{4} \\ -1-w^{2}-w^{4}\end{array}\right]$

$=\left[\begin{array}{c}-w\left(1+w+w^{2}\right) \\ -1-w^{2}-w^{3} w \\ -1-w^{2}-w^{3} w\end{array}\right]$

$=\left[\begin{array}{c}-w \times 0 \\ -1-w^{2}-w \\ -1-w^{2}-w\end{array}\right]$       $\left(\because 1+w+w^{2}=0\right.$ and $\left.w^{3}=1\right)$

$=\left[\begin{array}{c}0 \\ -0 \\ -0\end{array}\right]$

$=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$\therefore\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

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