If $A =\begin{pmatrix}\frac{-1+i\sqrt{3}}{2i}&\frac{-1-i\sqrt{3}}{2i}\\\frac{1+i\sqrt{3}}{2i}&\frac{1-i\sqrt{3}}{2i}\end{pmatrix}, i = \sqrt{-1}$ and $f(x) = x^2 + 2$, then f(A) equals

$(2+i\sqrt{3})\begin{pmatrix}1&0\\0&1\end{pmatrix}$

$∵ω=\frac{-1+i\sqrt{3}}{2}$ and $ω^2=\frac{-1-i\sqrt{3}}{2}$

Also, $ω^3=1$ and $ω+ω^2=-1$

Then $A =\begin{pmatrix}\frac{ω}{i}&\frac{ω^2}{2i}\\-\frac{ω^2}{i}&-\frac{ω}{i}\end{pmatrix}=\frac{ω}{i}\begin{pmatrix}1&ω\\-ω&-1\end{pmatrix}$

$∴A^2=\frac{ω^2}{i^2}\begin{pmatrix}1&ω\\-ω&-1\end{pmatrix}\begin{pmatrix}1&ω\\-ω&-1\end{pmatrix}$

$-ω^2\begin{pmatrix}1-ω^2&0\\0&1-ω^2\end{pmatrix}=\begin{pmatrix}-ω^2+ω^4&0\\0&-ω^2+ω^4\end{pmatrix}$

$=\begin{pmatrix}-ω^2+ω&0\\0&-ω^2+ω\end{pmatrix}+\begin{pmatrix}2&0\\0&2\end{pmatrix}$

$=\begin{pmatrix}-ω^2+ω+2&0\\0&-ω^2+ω+2\end{pmatrix}$

$=(-ω^2+ω+2)\begin{pmatrix}1&0\\0&1\end{pmatrix}$

$=(3+2ω)\begin{pmatrix}1&0\\0&1\end{pmatrix}=\left(3+2\left(\frac{-1+i\sqrt{3}}{2}\right)\right)\begin{pmatrix}1&0\\0&1\end{pmatrix}$

$=(2+i\sqrt{3})\begin{pmatrix}1&0\\0&1\end{pmatrix}$