Question

Let $A=\left[\begin{array}{cc}1& \frac{-1-i\sqrt{3}}{2}\\ \frac{-1+i\sqrt{3}}{2}& 1\end{array}\right]$. Then, $A^{100}=$

• A. $2^{100}A$
• B. $2^{99}A$
• C. $2^{98}A$
• D. $A$
• E. $A^2$

Answer 1

The correct option is B $2^{99}A$

$A=\left[\begin{array}{cc}1& \frac{-1-i\sqrt{3}}{2}\\ \frac{-1+i\sqrt{3}}{2}& 1\end{array}\right]$
Let $\frac{-1+i\sqrt{3}}{2}=\omega$
$\Rightarrow {\omega }^2=\frac{{(-1+i\sqrt{3})}^2}{{\left(2\right)}^2}=\frac{1-2i\sqrt{3}-3}{4}$

$\Rightarrow {\omega }^2=\frac{-2-2i\sqrt{3}}{4}=\frac{-1-i\sqrt{3}}{2}$
$\therefore A=\left[\begin{array}{cc}1& {\omega }^2\\ \omega & 1\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}1& {\omega }^2\\ \omega & 1\end{array}\right]\left[\begin{array}{cc}1& {\omega }^2\\ \omega & 1\end{array}\right]=\left[\begin{array}{cc}2& {2\omega }^2\\ 2\omega & 2\end{array}\right]$
$[\because {\omega }^2=1]$
$=2\left[\begin{array}{cc}1& {\omega }^2\\ \omega & 1\end{array}\right]=2A$
$\Rightarrow A^3=A^2.A=2A.A=4A$
$\therefore A^n=2^{n-1}A$
$\Rightarrow A^{100}=2^{100-1}A=2^{99}A$