Animated Solution for Mathematics - Matrices and Determinants: 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

The quadratic equation $x^2+x+1=0$ serves as the foundation for the complex cube roots of unity. The roots of this equation are $\omega$ and ${\omega }^2$, where $\omega =e^{i\frac{2\pi }{3}}$.

These roots represent the vertices of an equilateral triangle inscribed in the unit circle. Recognizing these values is the key to unlocking the structure of the given matrix.

We are given the matrix $A$ defined as:

By substituting $\alpha =\omega$ and utilizing the properties ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$, we simplify the term ${\alpha }^4$ to ${\omega }^4={\omega }^3\cdot \omega =\omega$. The matrix simplifies to:

To compute $A^{31}$, we must identify the cyclic behavior of the matrix powers. We begin by calculating $A^2$:

Performing the row-by-column multiplication, the resulting matrix is:

We now determine the period of the matrix by calculating $A^4$:

The matrix $A$ follows a cycle of four, where $A^4=I_3$.

With the period established, we reduce the exponent $31$ using the division algorithm: $31=4\times 7+3$. Consequently:

Since $A^3=A^2\cdot A$, we calculate:

The final result is $A^{31}=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& {\omega }^2& \omega \\ 1& \omega & {\omega }^2\end{array}\right]$.