If `A=[[alpha,beta],[gamma,alpha]]`is such that `A^2=I`, then

Options

A. `1+alpha^2+betagamma=0`
B. `1-alpha^2+betagamma=0`
C. `1-alpha^2-betagamma=0`
D. `1+alpha^2-betagamma=0`

Explanation

To solve the problem, we need to find the conditions on the elements of the matrix \( A = \begin{pmatrix} \alpha & \beta \\ \gamma & \alpha \end{pmatrix} \) given that \( A^2 = I \), where \( I \) is the identity matrix.

### Step-by-step Solution:

1. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} \alpha & \beta \\ \gamma & \alpha \end{pmatrix} \cdot \begin{pmatrix} \alpha & \beta \\ \gamma & \alpha \end{pmatrix} \]
Using matrix multiplication:
\[ A^2 = \begin{pmatrix} \alpha^2 + \beta \gamma & \alpha \beta + \beta \alpha \\ \gamma \alpha + \alpha \gamma & \gamma \beta + \alpha^2 \end{pmatrix} \]
Simplifying the terms:
\[ A^2 = \begin{pmatrix} \alpha^2 + \beta \gamma & 2\alpha \beta \\ 2\gamma \alpha & \gamma \beta + \alpha^2 \end{pmatrix} \]

2. **Set \( A^2 \) equal to the identity matrix \( I \)**:
The identity matrix is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Therefore, we have:
\[ \begin{pmatrix} \alpha^2 + \beta \gamma & 2\alpha \beta \\ 2\gamma \alpha & \gamma \beta + \alpha^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

3. **Set up the equations**:
From the equality of the matrices, we get the following equations:
- \( \alpha^2 + \beta \gamma = 1 \) (1)
- \( 2\alpha \beta = 0 \) (2)
- \( 2\gamma \alpha = 0 \) (3)
- \( \gamma \beta + \alpha^2 = 1 \) (4)

4. **Solve the equations**:
From equation (2) \( 2\alpha \beta = 0 \):
- This implies either \( \alpha = 0 \) or \( \beta = 0 \).

From equation (3) \( 2\gamma \alpha = 0 \):
- This implies either \( \gamma = 0 \) or \( \alpha = 0 \).

Now we consider two cases:

**Case 1**: If \( \alpha = 0 \):
- From (1): \( 0 + \beta \gamma = 1 \) → \( \beta \gamma = 1 \)
- From (4): \( \gamma \beta + 0 = 1 \) → \( \gamma \beta = 1 \) (consistent).

**Case 2**: If \( \beta = 0 \):
- From (1): \( \alpha^2 + 0 = 1 \) → \( \alpha^2 = 1 \) → \( \alpha = 1 \) or \( \alpha = -1 \).
- From (4): \( 0 + \alpha^2 = 1 \) → \( \alpha^2 = 1 \) (consistent).

5. **Conclusion**:
The conditions derived are:
- If \( \alpha = 0 \), then \( \beta \gamma = 1 \).
- If \( \beta = 0 \), then \( \alpha = 1 \) or \( \alpha = -1 \).