Solve the following
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:

1. (1) $A$

2. (2) $I_{3}$

3. (3) $A^{2}$

4. (4) $A^{3}$

Correct Option: , 4

Solution:

Solution of $x^{2}+x+1=0$ is $\omega, \omega^{2}$

So, $\alpha=\omega$ and

$\omega^{4}=\omega^{3} \cdot \omega=1 \cdot \omega=\omega$

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

$\Rightarrow A^{4}=I$

$\Rightarrow A^{30}=A^{28} \times A^{3}=A^{3}$