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:
Correct Option: , 4
Solution of $x^{2}+x+1=0$ is $\omega, \omega^{2}$
So, $\alpha=\omega$ and
$\omega^{4}=\omega^{3} \cdot \omega=1 \cdot \omega=\omega$
$A^{2}=\frac{1}{3}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$
$\Rightarrow A^{4}=I$
$\Rightarrow A^{30}=A^{28} \times A^{3}=A^{3}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.