Question

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

• A. $\left(\frac{5-i\sqrt{3}}{2}\right)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• B. $\left(\frac{3-i\sqrt{3}}{2}\right)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• D. $(2+i\sqrt{3})\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Answer 1

The correct option is D $(2+i\sqrt{3})\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

From the given matrix, we have

$A=\left[\begin{array}{cc}\frac{\omega }{i}& \frac{{\omega }^2}{i}\\ -\frac{{\omega }^2}{i}& -\frac{\omega }{i}\end{array}\right]=\frac{\omega }{i}\left[\begin{array}{cc}1& \omega \\ -\omega & -1\end{array}\right]$
Thus $A^2={\omega }^2\left[\begin{array}{cc}1-{\omega }^2& 0\\ 0& 1-{\omega }^2\end{array}\right]$
$=-\left[\begin{array}{cc}-{\omega }^2+{\omega }^4& 0\\ 0& -{\omega }^2+\omega \end{array}\right]$
$=\left[\begin{array}{cc}-{\omega }^2+\omega & 0\\ 0& -{\omega }^2+\omega \end{array}\right]$ ....$\text{Since}f\left(x\right)=x^2+2$
Therefore, $f\left(A\right)=A^2+2=\left[\begin{array}{cc}-{\omega }^2+\omega & 0\\ 0& -{\omega }^2+\omega \end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$
$-[{\omega }^2+\omega +2]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=(3+2\omega )\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=(2+i\sqrt{3})\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$