Let `alpha` be a root of the equation `x ^(2) - x+1=0,` and the matrix `A=[{:(1,1,1),(1, alpha , alpha ^(2)), (1, alpha ^(2), alpha ^(4)):}]` and matrix `B= [{:(1,-1, -1),(1, alpha, - alpha ^(2)),(-1, -alpha ^(2), - alpha ^(4)):}]` then the vlaue of |AB| is:

Options

A. 1
B. `-1`
C. 3
D. `-3`

Explanation

To solve the problem step by step, we need to find the determinant of the product of two matrices \( A \) and \( B \). We will first find the determinants of each matrix separately and then multiply them.

### Step 1: Identify the roots of the equation
The roots of the equation \( x^2 - x + 1 = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = 1 \):
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2} \]
Let \( \alpha = \frac{1 + i\sqrt{3}}{2} \) (one of the roots).

### Step 2: Define the matrices
The matrices are given as:
\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha^4 \end{pmatrix} \]
\[ B = \begin{pmatrix} 1 & -1 & -1 \\ 1 & \alpha & -\alpha^2 \\ -1 & -\alpha^2 & -\alpha^4 \end{pmatrix} \]

### Step 3: Calculate the determinant of matrix \( A \)
Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Substituting the values from matrix \( A \):
\[ |A| = 1(\alpha \cdot \alpha^4 - \alpha^2 \cdot 1) - 1(1 \cdot \alpha^4 - \alpha^2 \cdot 1) + 1(1 \cdot \alpha^2 - \alpha \cdot 1) \]
Calculating each term:
1. \( \alpha \cdot \alpha^4 - \alpha^2 \cdot 1 = \alpha^5 - \alpha^2 \)
2. \( 1 \cdot \alpha^4 - \alpha^2 \cdot 1 = \alpha^4 - \alpha^2 \)
3. \( 1 \cdot \alpha^2 - \alpha \cdot 1 = \alpha^2 - \alpha \)

Now substituting back:
\[ |A| = \alpha^5 - \alpha^2 - (\alpha^4 - \alpha^2) + (\alpha^2 - \alpha) \]
Simplifying:
\[ |A| = \alpha^5 - \alpha^4 + \alpha - \alpha \]
Using \( \alpha^2 = \alpha - 1 \) (from the original equation):
\[ |A| = \alpha^5 - \alpha^4 \]
Using \( \alpha^3 = \alpha^2 \cdot \alpha = (\alpha - 1) \cdot \alpha = \alpha^2 - \alpha = -1 \):
Thus, \( \alpha^5 = \alpha^2 \) and \( \alpha^4 = \alpha \):
\[ |A| = \alpha^2 - \alpha = -1 \]

### Step 4: Calculate the determinant of matrix \( B \)
Using a similar approach:
\[ |B| = 1(-\alpha^2 \cdot -\alpha^4 - (-1)(-1)) - (-1)(1 \cdot -\alpha^4 - (-\alpha^2)(-1)) + (-1)(1 \cdot -\alpha^2 - (-1)(1)) \]
Calculating each term:
1. \( -\alpha^2 \cdot -\alpha^4 - 1 = \alpha^6 - 1 = 1 - 1 = 0 \)
2. \( -(-1)(-\alpha^4 - \alpha^2) = \alpha^4 + \alpha^2 = \alpha + \alpha - 1 = 0 \)
3. \( -(-1)(-\alpha^2 - 1) = \alpha^2 + 1 = 0 \)

Thus, \( |B| = 0 \).

### Step 5: Calculate the determinant of the product \( |AB| \)
Using the property of determinants:
\[ |AB| = |A| \cdot |B| = (-1) \cdot 0 = 0 \]

### Final Answer
The value of \( |AB| \) is \( 0 \).