If $A=\left[\begin{array}{cc}\frac{-1+\mathrm{i}\sqrt{3}}{2\mathrm{i}}& \frac{-1-\mathrm{i}\sqrt{3}}{2\mathrm{i}}\\ \frac{1+\mathrm{i}\sqrt{3}}{2\mathrm{i}}& \frac{1-\mathrm{i}\sqrt{3}}{2\mathrm{i}}\end{array}\right], \mathrm{i}=\sqrt{-1}$ and $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+2$ then $f\left(A\right)$ equals to

$$\begin{gathered} \because \mathrm{\omega }=\frac{-1+\mathrm{i}\sqrt{3}}{2} \mathrm{and} {\mathrm{\omega }}^2=\frac{-1-\mathrm{i}\sqrt{3}}{2} \\ \mathrm{Also}, {\mathrm{\omega }}^3=1 \mathrm{and} \mathrm{\omega }+{\mathrm{\omega }}^2=-1 \\ \mathrm{Thus}, \mathrm{A}=\left[\begin{array}{cc}-\mathrm{i\omega }& -{\mathrm{i\omega }}^2\\ {\mathrm{i\omega }}^2& \mathrm{i\omega }\end{array}\right] \\ \therefore {\mathrm{A}}^2=\left[\begin{array}{cc}-\mathrm{i\omega }& -{\mathrm{i\omega }}^2\\ {\mathrm{i\omega }}^2& \mathrm{i\omega }\end{array}\right]\left[\begin{array}{cc}-\mathrm{i\omega }& -{\mathrm{i\omega }}^2\\ {\mathrm{i\omega }}^2& \mathrm{i\omega }\end{array}\right]=\left[\begin{array}{cc}-{\mathrm{\omega }}^2+\mathrm{\omega }& 0\\ 0& -{\mathrm{\omega }}^2+\mathrm{\omega }\end{array}\right] \\ \mathrm{Now}, \mathrm{f}\left(\mathrm{A}\right)={\mathrm{A}}^2+2\mathrm{I}=\left[\begin{array}{cc}-{\mathrm{\omega }}^2+\mathrm{\omega }& 0\\ 0& -{\mathrm{\omega }}^2+\mathrm{\omega }\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right] \\ =\left[\begin{array}{cc}-{\mathrm{\omega }}^2+\mathrm{\omega }+2& 0\\ 0& -{\mathrm{\omega }}^2+\mathrm{\omega }+2\end{array}\right] \\ =(-{\mathrm{\omega }}^2+\mathrm{\omega }+2)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=(2+\mathrm{i}\sqrt{3})\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$