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, $A^{100}$ is equal to

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

Answer 1

Answer: (B)

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