Question

If w is a complex cube root of unity, show that
$\left(\left[\begin{array}{ccc}1& w& w^2\\ w& w^2& 1\\ w^2& 1& w\end{array}\right]+\left[\begin{array}{ccc}w& w^2& 1\\ w^2& 1& w\\ w& w^2& 1\end{array}\right]\right)\left[\begin{array}{c}1\\ w\\ w^2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{LHS}=\left(\left[\begin{array}{ccc}1& w& w^2\\ w& w^2& 1\\ w^2& 1& w\end{array}\right]+\left[\begin{array}{ccc}w& w^2& 1\\ w^2& 1& w\\ w& w^2& 1\end{array}\right]\right)\left[\begin{array}{c}1\\ w\\ w^2\end{array}\right] \\ =\left[\begin{array}{ccc}1+w& w+w^2& w^2+1\\ w+w^2& w^2+1& 1+w\\ w^2+w& 1+w^2& w+1\end{array}\right]\left[\begin{array}{c}1\\ w\\ w^2\end{array}\right] \\ =\left[\begin{array}{ccc}-w^2& -1& -w\\ -1& -w& -w^2\\ -1& -w& -w^2\end{array}\right]\left[\begin{array}{c}1\\ w\\ w^2\end{array}\right]\left(\because 1+\mathrm{w}+{\mathrm{w}}^2=0\mathrm{and}{\mathrm{w}}^3=1\right) \\ =\left[\begin{array}{c}-w^2-w-w^3\\ -1-w^2-w^4\\ -1-w^2-w^4\end{array}\right] \\ =\left[\begin{array}{c}-w\left(1+w+w^2\right)\\ -1-w^2-w^{3}w\\ -1-w^2-w^{3}w\end{array}\right] \\ =\left[\begin{array}{c}-w\times 0\\ -1-w^2-w\\ -1-w^2-w\end{array}\right]\left(\because 1+\mathrm{w}+{\mathrm{w}}^2=0\mathrm{and}{\mathrm{w}}^3=1\right) \\ =\left[\begin{array}{c}0\\ -0\\ -0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \therefore \left(\left[\begin{array}{ccc}1& w& w^2\\ w& w^2& 1\\ w^2& 1& w\end{array}\right]+\left[\begin{array}{ccc}w& w^2& 1\\ w^2& 1& w\\ w& w^2& 1\end{array}\right]\right)\left[\begin{array}{c}1\\ w\\ w^2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$