If `A=[{:(alpha, beta),(gamma,-alpha):}]` is such that `A^(2)=I` then

Options

A. `1+alpha^(2)+betagamma=0`
B. `1-alpha^(2)+beta gamma=0`
C. `1-alpha^(2)-beta gamma=0`
D. `1+alpha^(-2)-beta gamma=0`

Answer

To solve the problem, we need to find the relationship between the elements of the matrix \( A \) given that \( A^2 = I \), where \( I \) is the identity matrix.

Given:
\[ A = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)
We need to compute \( A^2 = A \times A \):
\[ A^2 = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \times \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \]

### Step 2: Perform the matrix multiplication
Calculating the elements of the resulting matrix:
- The element at (1,1):
\[ \alpha \cdot \alpha + \beta \cdot \gamma = \alpha^2 + \beta\gamma \]
- The element at (1,2):
\[ \alpha \cdot \beta + \beta \cdot (-\alpha) = \alpha\beta - \alpha\beta = 0 \]
- The element at (2,1):
\[ \gamma \cdot \alpha + (-\alpha) \cdot \gamma = \gamma\alpha - \alpha\gamma = 0 \]
- The element at (2,2):
\[ \gamma \cdot \beta + (-\alpha) \cdot (-\alpha) = \gamma\beta + \alpha^2 \]

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

### Step 3: Set \( A^2 \) equal to the identity matrix
Since \( A^2 = I \), we have:
\[ \begin{pmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \gamma\beta + \alpha^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Equate the corresponding elements
From the equality of the matrices, we get two equations:
1. \( \alpha^2 + \beta\gamma = 1 \)
2. \( \gamma\beta + \alpha^2 = 1 \)

### Step 5: Simplify the equations
Both equations are actually the same, so we can simplify it to:
\[ \alpha^2 + \beta\gamma = 1 \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ 1 - \alpha^2 - \beta\gamma = 0 \]

### Conclusion
Thus, the final equation is:
\[ 1 - \alpha^2 - \beta\gamma = 0 \]