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

We have the quadratic equation $$x^{2}+x+1=0$$. Since the constant term and the coefficient of the middle term are both $$1$$, its roots are the non-real cube roots of unity. Hence any root $$\alpha$$ of this equation satisfies the two standard relations

$$\alpha^{2}+\alpha+1=0 \quad\Longrightarrow\quad \alpha^{3}=1,\qquad\text{and therefore}\qquad\alpha^{4}=\alpha.$$

The given matrix is written more compactly by extracting the common scalar factor:

$$A=\dfrac1{\sqrt3}\,M,\quad\text{where}\quad M=\begin{bmatrix} 1&1&1\\ 1&\alpha&\alpha^{2}\\ 1&\alpha^{2}&\alpha^{4} \end{bmatrix} =\begin{bmatrix} 1&1&1\\ 1&\alpha&\alpha^{2}\\ 1&\alpha^{2}&\alpha \end{bmatrix}.$$

To obtain powers of $$A$$ we first compute $$M^{2}$$ entry by entry. The general rule used is $$(M^{2})_{ij}=R_{i}\cdot C_{j},$$ where $$R_{i}$$ is the $$i^{\text{th}}$$ row and $$C_{j}$$ the $$j^{\text{th}}$$ column of $$M$$.

First row of $$M^{2}$$

$$\begin{aligned} (M^{2})_{11}&=1+1+1=3,\\ (M^{2})_{12}&=1+\alpha+\alpha^{2}=0,\\ (M^{2})_{13}&=1+\alpha^{2}+\alpha=0. \end{aligned}$$

Second row of $$M^{2}$$

$$\begin{aligned} (M^{2})_{21}&=1+\alpha+\alpha^{2}=0,\\ (M^{2})_{22}&=1+\alpha^{2}+\alpha=0,\\ (M^{2})_{23}&=1+\alpha\alpha^{2}+\alpha^{2}\alpha=1+1+1=3. \end{aligned}$$

Third row of $$M^{2}$$

$$\begin{aligned} (M^{2})_{31}&=1+\alpha^{2}+\alpha=0,\\ (M^{2})_{32}&=1+\alpha^{2}\alpha+\alpha\alpha^{2}=1+1+1=3,\\ (M^{2})_{33}&=1+\alpha+\alpha^{2}=0. \end{aligned}$$

Collecting all entries gives

$$M^{2}= \begin{bmatrix} 3&0&0\\ 0&0&3\\ 0&3&0 \end{bmatrix}.$$

Using $$A=\dfrac1{\sqrt3}M$$ we now square $$A$$:

$$A^{2}=\left(\dfrac1{\sqrt3}M\right)^{2} =\dfrac1{3}\,M^{2} =\dfrac1{3}\begin{bmatrix} 3&0&0\\ 0&0&3\\ 0&3&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix}.$$

The matrix obtained is merely a permutation matrix that keeps the first component intact and swaps the second and third components. Denote it by $$P$$, viz.

$$P=\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix},\qquad\text{so that}\qquad A^{2}=P.$$

A permutation matrix squared either remains the same or returns to the identity; here, swapping twice restores the original order, so

$$P^{2}=I_{3}\quad\Longrightarrow\quad A^{4}=(A^{2})^{2}=P^{2}=I_{3}.$$

The equality $$A^{4}=I_{3}$$ shows that the order of $$A$$ divides $$4$$. To find $$A^{31}$$ we reduce the exponent modulo $$4$$:

$$31=4\times7+3\;\Longrightarrow\;A^{31}=A^{(4\times7)+3}=(A^{4})^{7}\,A^{3}=I_{3}^{\,7}\,A^{3}=A^{3}.$$

Therefore $$A^{31}=A^{3}$$, which matches Option A.

Hence, the correct answer is Option A.