Question

Let be $\alpha$ root of the equation $x^2+x+1=0$ and the matrix $A=$ $\begin{array}{l}\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& {\alpha }^2& {\alpha }^4\end{array}\right]\end{array}$ 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

$A^3$

Explanation for the correct option:

Finding the value of the matrix $A^{31}$ :

The equation $x^2+x+1=0$ has complex roots of unity $1,\omega \text{or}{\omega }^2$

$\Rightarrow 1+\omega +{\omega }^2=0$

$\begin{array}{l}A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& {\alpha }^2& {\alpha }^4\end{array}\right]\end{array}$here $\alpha$ is a root.

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

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

$\Rightarrow \begin{array}{l}=\frac{1}{3}\left[\begin{array}{ccc}3& 0& 0\\ 0& 0& 3\\ 0& 3& 0\end{array}\right]\end{array}$ $\left[\because {\omega }^3=1,1+\omega +{\omega }^2=0\right]$

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

$\Rightarrow \begin{array}{l}{A}^4=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]\end{array}$ [multiplying $A^2$on both sides]

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

$$\begin{gathered} \Rightarrow A^4=I \\ \Rightarrow {\left(A^4\right)}^7={\left(I\right)}^7 \\ \Rightarrow A^{28}{A}^3=I{A}^3 \\ \Rightarrow A^{31}=A^3 \end{gathered}$$

Hence, option (C) is correct.