Let ` alpha ` be a root of the equation ` x ^(2) + x + 1 = 0 ` and the matrix ` A = ( 1 ) /(sqrt3) [{:( 1,,1,,1),( 1,, alpha ,, alpha ^(2)), ( 1 ,, alpha ^(2),, alpha ^(4)):}] `
then the matrix ` A ^( 31 ) ` is equal to :

Options

A. ` I _ 3 `
B. ` A ^(2) `
C. ` A `
D. ` A ^( 3 ) `

Answer

To solve the problem step by step, we will first identify the roots of the given quadratic equation, then we will analyze the matrix \( A \), and finally compute \( A^{31} \).

### Step 1: Find the roots of the equation \( x^2 + x + 1 = 0 \)

The roots of the equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, the roots are complex:
\[ x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Let \( \alpha = \frac{-1 + i\sqrt{3}}{2} \) and \( \alpha^2 = \frac{-1 - i\sqrt{3}}{2} \).

### Step 2: Define the matrix \( A \)

The matrix \( A \) is given by:
\[ A = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha^4 \end{pmatrix} \]
Since \( \alpha^3 = 1 \) (as \( \alpha \) is a cube root of unity), we can replace \( \alpha^4 \) with \( \alpha \).

Thus, the matrix \( A \) simplifies to:
\[ A = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{pmatrix} \]

### Step 3: Compute \( A^2 \)

To find \( A^2 \), we compute:
\[ A^2 = A \cdot A = \left(\frac{1}{\sqrt{3}}\right)^2 \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{pmatrix} \]

Calculating the product:
\[ A^2 = \frac{1}{3} \begin{pmatrix} 3 & 3 & 3 \\ 3 & 1 + \alpha + \alpha^2 & \alpha + \alpha^2 + \alpha^3 \\ 3 & \alpha^2 + \alpha + \alpha^4 & \alpha^2 + \alpha^3 + \alpha \end{pmatrix} \]
Using \( 1 + \alpha + \alpha^2 = 0 \) and \( \alpha^3 = 1 \):
\[ A^2 = \frac{1}{3} \begin{pmatrix} 3 & 3 & 3 \\ 3 & 0 & 0 \\ 3 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \]

### Step 4: Compute \( A^3 \)

Now, we compute \( A^3 = A^2 \cdot A \):
\[ A^3 = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \cdot A \]
Calculating this product gives us:
\[ A^3 = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \cdot \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{pmatrix} \]
This results in:
\[ A^3 = \frac{1}{\sqrt{3}} \begin{pmatrix} 3 & 3 & 3 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{pmatrix} \]

### Step 5: Compute \( A^{31} \)

Since \( A^3 \) is a constant matrix, we can express \( A^{31} \) as:
\[ A^{31} = A^{30} \cdot A = (A^3)^{10} \cdot A \]
Since \( A^3 \) is a constant matrix, we can conclude:
\[ A^{31} = A^3 \]

### Final Result

Thus, the final result is:
\[ A^{31} = A^3 \]