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

The correct option is C $A^3$

The roots of equation $x^2+x+1=0$ are complex cube roots of unity.
$\therefore \alpha =\omega$ or ${\omega }^2$

$A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& {\alpha }^2& {\alpha }^4\end{array}\right]=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]$

$A^2=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]$

$\Rightarrow A^2=\frac{1}{3}\left[\begin{array}{ccc}3& 0& 0\\ 0& 0& 3\\ 0& 3& 0\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$

$A^4=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$

$\Rightarrow A^4=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I{A}^4=I\Rightarrow A^{28}=I$

Therefore, we get
$A^{31}=A^{28}{A}^3\Rightarrow A^{31}=I{A}^3\Rightarrow A^{31}=A^3$