If for the matrix , `A = [(1,-alpha),(alpha, beta)], A A^T = I_2`, then the value of `alpha^4 + beta^4` is:

Options

A. `4`
B. `2`
C. `3`
D. `1`

Answer

To solve the problem, we need to find the value of \( \alpha^4 + \beta^4 \) given that the matrix \( A = \begin{pmatrix} 1 & -\alpha \\ \alpha & \beta \end{pmatrix} \) satisfies the condition \( A A^T = I_2 \), where \( I_2 \) is the identity matrix of size 2.

### Step-by-step Solution:

1. **Find the Transpose of Matrix A**:
The transpose of matrix \( A \) is given by:
\[ A^T = \begin{pmatrix} 1 & \alpha \\ -\alpha & \beta \end{pmatrix} \]

2. **Multiply A and A^T**:
Now we compute the product \( A A^T \):
\[ A A^T = \begin{pmatrix} 1 & -\alpha \\ \alpha & \beta \end{pmatrix} \begin{pmatrix} 1 & \alpha \\ -\alpha & \beta \end{pmatrix} \]
Performing the multiplication:
- The element at (1,1) is \( 1 \cdot 1 + (-\alpha)(-\alpha) = 1 + \alpha^2 \)
- The element at (1,2) is \( 1 \cdot \alpha + (-\alpha) \cdot \beta = \alpha - \alpha\beta \)
- The element at (2,1) is \( \alpha \cdot 1 + \beta \cdot (-\alpha) = \alpha - \alpha\beta \)
- The element at (2,2) is \( \alpha \cdot \alpha + \beta \cdot \beta = \alpha^2 + \beta^2 \)

Thus, we have:
\[ A A^T = \begin{pmatrix} 1 + \alpha^2 & \alpha - \alpha\beta \\ \alpha - \alpha\beta & \alpha^2 + \beta^2 \end{pmatrix} \]

3. **Set the Product Equal to the Identity Matrix**:
Since \( A A^T = I_2 \), we equate it to the identity matrix:
\[ \begin{pmatrix} 1 + \alpha^2 & \alpha - \alpha\beta \\ \alpha - \alpha\beta & \alpha^2 + \beta^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

4. **Formulate the Equations**:
From the above equality, we can form the following equations:
- From the (1,1) entry: \( 1 + \alpha^2 = 1 \) → \( \alpha^2 = 0 \)
- From the (1,2) entry: \( \alpha - \alpha\beta = 0 \)
- From the (2,2) entry: \( \alpha^2 + \beta^2 = 1 \)

5. **Solve for Alpha and Beta**:
From \( \alpha^2 = 0 \), we find:
\[ \alpha = 0 \]
Substituting \( \alpha = 0 \) into \( \alpha - \alpha\beta = 0 \) gives no new information. Now substituting \( \alpha = 0 \) into \( \alpha^2 + \beta^2 = 1 \):
\[ 0 + \beta^2 = 1 \implies \beta^2 = 1 \implies \beta = \pm 1 \]

6. **Calculate \( \alpha^4 + \beta^4 \)**:
Now we compute \( \alpha^4 + \beta^4 \):
\[ \alpha^4 + \beta^4 = 0^4 + 1^4 = 0 + 1 = 1 \]

### Final Answer:
Thus, the value of \( \alpha^4 + \beta^4 \) is \( \boxed{1} \).