Question

Let $\alpha$ be a root of the equation $x^2+x+1=0$ and the matrix $A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& {\alpha }^2& {\alpha }^4\end{array}\right]$, then the matrix $A^{31}$ is equal to :

• A. $A$
• B. $A^2$
• C. $A^3$
• D. $I_3$

Answer 1

Answer: (C)

$x^2+x+1=0$
$\alpha =\omega$
${\alpha }^2={\omega }^2$
$A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]$
$A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$
$\Rightarrow A^4=A^2.A^2=I_3$
$A^{31}=A^{28}.A^3=A^{3.}$