Given that $A=\begin{bmatrix} 1 &0 &0 \\ i &\frac{-1+i\sqrt{3}}{2} &0 \\ 0& 1+2i & \frac{-1-i\sqrt{3}}{2} \end{bmatrix}$
We know that cube root of unity
$Z^3=1$
We get $Z=1,\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$
$Z=1,w,w^{2}$
and we know that $1+w+w^{2}=0,w^{3}=1$
$A^{2}=\begin{bmatrix} 1 &0 &0 \\ i &\frac{-1-i\sqrt{3}}{2} &0 \\ 0& 1+2i & \frac{-1+i\sqrt{3}}{2} \end{bmatrix}$
$A^{3}=\begin{bmatrix} 1 &0 &0 \\ i &\frac{-1+i\sqrt{3}}{2} &0 \\ 0& 1+2i & \frac{-1-i\sqrt{3}}{2} \end{bmatrix}$
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$A^{101}=\begin{bmatrix} 1 &0 &0 \\ i &\frac{-1+i\sqrt{3}}{2} &0 \\ 0& 1+2i & \frac{-1-i\sqrt{3}}{2} \end{bmatrix}$
$A^{102}=\begin{bmatrix} 1 &0 &0 \\ i &\frac{-1-i\sqrt{3}}{2} &0 \\ 0& 1+2i & \frac{-1+i\sqrt{3}}{2} \end{bmatrix}$
$(OR)$
$A=\begin{bmatrix} 1 &0 &0 \\ i &w&0 \\ 0& 1+2i & w^{2} \end{bmatrix}$
$A^{2}=\begin{bmatrix} 1 &0 &0 \\ i &w^{2} &0 \\ 0& 1+2i &w \end{bmatrix}$
$A^{3}=\begin{bmatrix} 1 &0 &0 \\ i &w&0 \\ 0& 1+2i & w^{2} \end{bmatrix}$
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$A^{101}=\begin{bmatrix} 1 &0 &0 \\ i &w&0 \\ 0& 1+2i & w^{2} \end{bmatrix}$
$A^{102}=\begin{bmatrix} 1 &0 &0 \\ i &w^{2} &0 \\ 0& 1+2i &w \end{bmatrix}$
Trace of the matrix$(T_{r})=$sum of the leading diagonal elements.
$T_{r}=1+w^{2}+w=0$
