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`.

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 calculate \( A^2 \) by multiplying \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \cdot \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \]

### Step 2: Perform the multiplication
Using the matrix multiplication rules:
1. The element in the first row, first column:
\[ \alpha \cdot \alpha + \beta \cdot \gamma = \alpha^2 + \beta \gamma \]

2. The element in the first row, second column:
\[ \alpha \cdot \beta + \beta \cdot (-\alpha) = \alpha \beta - \beta \alpha = 0 \]

3. The element in the second row, first column:
\[ \gamma \cdot \alpha + (-\alpha) \cdot \gamma = \gamma \alpha - \alpha \gamma = 0 \]

4. The element in the second row, second column:
\[ \gamma \cdot \beta + (-\alpha) \cdot (-\alpha) = \gamma \beta + \alpha^2 \]

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

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

Both equations are actually the same, so we can focus on one:
\[ \alpha^2 + \beta \gamma = 1 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 1 - \alpha^2 - \beta \gamma = 0 \]

### Conclusion
Thus, we find that:
\[ 1 - \alpha^2 - \beta \gamma = 0 \]

This corresponds to option 3 from the given choices.

### Final Answer
The correct option is:
\[ 1 - \alpha^2 - \beta \gamma = 0 \]

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