Given $A=\left[\begin{matrix}\dfrac{-1+i\sqrt{3}}{2i}&\dfrac{-1-i\sqrt{3}}{2i}\\\dfrac{1+i\sqrt{3}}{2i}&\dfrac{1-i\sqrt{3}}{2i}\end{matrix}\right]$ where $i=\sqrt{-1}$.
Also Given$f(x)=x^2+2.$
Find $f(A)$
My approach:
A can be written as
$A=i\left[\begin{matrix}w&w^2\\w^2&w\end{matrix}\right]$ Here $1+w+w^2=0$ i.e $w,w^2 \in {cube \space roots\space of \space \space unity }$
$\therefore \space f(A)=-((\left[\begin{matrix}w&w^2\\w^2&w\end{matrix}\right])^2-2)$
$=2-\left[\begin{matrix}-1&2\\2&-1\end{matrix}\right]$
Another approach:
$A=\left[\begin{matrix}\dfrac{-1+i\sqrt{3}}{2i}&\dfrac{-1-i\sqrt{3}}{2i}\\\dfrac{1+i\sqrt{3}}{2i}&\dfrac{1-i\sqrt{3}}{2i}\end{matrix}\right]$
$=i\left[\begin{matrix} cos(\dfrac{2\pi}{3})+isin(\dfrac{2\pi}{3})&cos(\dfrac{2\pi}{3})-isin(\dfrac{2\pi}{3})\\-cos(\dfrac{2\pi}{3})+isin(\dfrac{2\pi}{3})&cos(\dfrac{2\pi}{3})-isin(\dfrac{2\pi}{3})\end{matrix}\right] $
Now using DeMoivres' Theorem I was thinking what to be done. Let me think...
Now the options given are:
a. $\left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
b. $\dfrac{3-i\sqrt{3}}{2} \cdot \left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
c. $\dfrac{5-i\sqrt{3}}{2} \cdot \left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
d. $(2+i\sqrt{3}) \cdot \left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
But my answer does not come close to any of the given options. Could someone selflessly help me out please!
Your expression for $A$ is not correct. You have $$ A=-i\begin{bmatrix} w&w^2\\ -w^2&-w\end{bmatrix}=-iw\begin{bmatrix} 1&w\\-w&-1\end{bmatrix} $$ Then $$ A^2+2I=-w^2\begin{bmatrix} 1-w^2&0\\0&1-w^2\end{bmatrix}+2I=\begin{bmatrix} w-w^2+2&0\\0&w-w^2+2\end{bmatrix}=(2+i\sqrt 3)I, $$ since $$ w-w^2+2=-\frac12+\frac i2\sqrt3+\frac12+\frac i2\sqrt3+2=2+i\sqrt3 $$