Question

Let $\alpha$ be a root of the equation $x^{2}+ x + 1 =0$ and the matrix $A = \frac{1}{\sqrt{3}}\begin{bmatrix}1&1&1\\ 1&\alpha&\alpha^{2}\\ 1&\alpha^{2}&\alpha^{4}\end{bmatrix}$, 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} $
$A = \frac{1}{\sqrt{3}}\begin{bmatrix}1&1&1\\ 1&\omega&\omega ^{2}\\ 1&\omega^{2} &\omega \end{bmatrix}$
$A^{2}= \begin{bmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{bmatrix}$
$\Rightarrow A^{4} = A^{2} . A^{2} = I_{3}$
$A^{31} = A^{28}. A^{3} = A^{3.}$