If `A=[(alpha,beta),(gamma,-alpha)]` is such that `A^2=I`, then the value of `1-alpha^2-beta gamma` is:

Options

A. `-1`
B. 0
C. 1
D. Can't be calculated

Answer

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

### Step-by-Step Solution:

1. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \cdot \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \]

2. **Perform the matrix multiplication**:
- First row, first column:
\[ \alpha \cdot \alpha + \beta \cdot \gamma = \alpha^2 + \beta \gamma \]
- First row, second column:
\[ \alpha \cdot \beta + \beta \cdot (-\alpha) = \alpha \beta - \beta \alpha = 0 \]
- Second row, first column:
\[ \gamma \cdot \alpha + (-\alpha) \cdot \gamma = \gamma \alpha - \alpha \gamma = 0 \]
- Second row, second column:
\[ \gamma \cdot \beta + (-\alpha) \cdot (-\alpha) = \gamma \beta + \alpha^2 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \alpha^2 + \gamma \beta \end{pmatrix} \]

3. **Set \( A^2 \) equal to the identity matrix \( I \)**:
\[ A^2 = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

4. **Equate the corresponding elements**:
From the first row, first column:
\[ \alpha^2 + \beta \gamma = 1 \]
From the second row, second column:
\[ \alpha^2 + \gamma \beta = 1 \]

Since both equations are the same, we can use either one.

5. **Rearranging the equation**:
\[ \alpha^2 + \beta \gamma = 1 \implies 1 - \alpha^2 - \beta \gamma = 0 \]

6. **Final result**:
Therefore, the value of \( 1 - \alpha^2 - \beta \gamma \) is:
\[ \boxed{0} \]