If `[(alpha, beta),(gamma,-alpha)]` is to be square root of the two rowed unit matrix, then `alpha, beta` and `gamma` satisfy the relation

Options

A. `1+alpha^(2)+beta gamma=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 relation that the variables \( \alpha \), \( \beta \), and \( \gamma \) satisfy given that the matrix \( \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \) is the square root of the two-row unit matrix \( I_2 \).

### Step-by-Step Solution:

1. **Define the Matrices**:
Let \( A = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \).
We know that \( A^2 = I_2 \), where \( I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

2. **Square the Matrix**:
We calculate \( A^2 \):
\[ A^2 = \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ \gamma & -\alpha \end{pmatrix} \]

3. **Perform the Matrix Multiplication**:
- The element at (1,1):
\[ \alpha \cdot \alpha + \beta \cdot \gamma = \alpha^2 + \beta \gamma \]
- The element at (1,2):
\[ \alpha \cdot \beta + \beta \cdot (-\alpha) = \alpha \beta - \beta \alpha = 0 \]
- 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:
\[ A^2 = \begin{pmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \gamma \beta + \alpha^2 \end{pmatrix} \]

4. **Set \( A^2 \) Equal to \( I_2 \)**:
Since \( A^2 = I_2 \), we equate:
\[ \begin{pmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \gamma \beta + \alpha^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

5. **Extract the Equations**:
From the equality of the matrices, we get two equations:
- From the (1,1) entry:
\[ \alpha^2 + \beta \gamma = 1 \quad \text{(1)} \]
- From the (2,2) entry:
\[ \gamma \beta + \alpha^2 = 1 \quad \text{(2)} \]

6. **Combine the Equations**:
Both equations (1) and (2) are essentially the same, thus confirming:
\[ \alpha^2 + \beta \gamma = 1 \]

### Final Relation:
The relation that \( \alpha \), \( \beta \), and \( \gamma \) satisfy is:
\[ \alpha^2 + \beta \gamma = 1 \]