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, we will follow these steps:

### Step 1: Find the roots of the equation
The equation given is \( x^2 + x + 1 = 0 \). We can find the roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Thus, the roots are:
\[ x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
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 as:
\[ A = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha^4 \end{pmatrix} \]
Since \( \alpha^3 = 1 \) (as \( \alpha \) is a root of unity), we can simplify \( \alpha^4 \) to \( \alpha \).

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

### Step 3: Calculate \( A^2 \)
To find \( A^{31} \), we first need to calculate \( A^2 \):
\[ A^2 = A \cdot A = \left(\frac{1}{\sqrt{3}}\right)^2 \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{pmatrix} \]
Calculating the product, we find:
- The first row will be \( (1 + 1 + 1, 1 + \alpha + \alpha^2, 1 + \alpha^2 + \alpha) = (3, 0, 0) \) because \( 1 + \alpha + \alpha^2 = 0 \).
- The second row will be \( (1 + \alpha + \alpha^2, \alpha + \alpha^2 + \alpha^3, \alpha^2 + \alpha^3 + \alpha) = (0, 1, 0) \).
- The third row will be \( (1 + \alpha^2 + \alpha, \alpha^2 + \alpha + \alpha^3, \alpha + \alpha^3 + \alpha^2) = (0, 0, 1) \).

Thus, we have:
\[ A^2 = \frac{1}{3} \begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \end{pmatrix} \]

### Step 4: Calculate higher powers of \( A \)
Since \( A^2 \) is a diagonal matrix, we can easily compute higher powers:
\[ A^4 = (A^2)^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{9} & 0 \\ 0 & 0 & \frac{1}{9} \end{pmatrix} \]
Continuing this process, we find that:
\[ A^{2n} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3^n} & 0 \\ 0 & 0 & \frac{1}{3^n} \end{pmatrix} \]
and
\[ A^{31} = A^{30} \cdot A = A^{2 \cdot 15} \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3^{15}} & 0 \\ 0 & 0 & \frac{1}{3^{15}} \end{pmatrix} \cdot A \]

### Step 5: Final Result
Thus, we find:
\[ A^{31} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{3^{15}} & 0 \\ 0 & 0 & \frac{1}{3^{15}} \end{pmatrix} \cdot \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha \end{pmatrix} \]

This results in the final matrix \( A^{31} \).