If $A=\begin{bmatrix}\alpha&\beta\ \gamma&-\alpha\end{bmatrix}$ is such that $A^{2}=I$, then

Question

Accepted Answer

To determine why Option 3 is the correct answer, we need to verify the condition $ A^2 = I $ for the matrix $ A = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix} $.

### Step-by-step Explanation:

1. **Calculate $ A^2 $:**

$$
   A^2 = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix}
   $$

Performing the matrix multiplication:

$$
   A^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta - \alpha\beta \ \gamma\alpha - \gamma\alpha & \gamma\beta + \alpha^2 \end{bmatrix}
   $$

Simplifying the terms:

$$
   A^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \ 0 & \alpha^2 + \beta\gamma \end{bmatrix}
   $$

2. **Set $ A^2 = I $:**

The identity matrix $ I $ is:

$$
   I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}
   $$

For $ A^2 $ to equal $ I $, we must have:

$$
   \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \ 0 & \alpha^2 + \beta\gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}
   $$

This implies:

$$
   \alpha^2 + \beta\gamma = 1
   $$

3. **Rearrange the Equation:**

Rearranging the equation $ \alpha^2 + \beta\gamma = 1 $ gives:

$$
   1 - \alpha^2 - \beta\gamma = 0
   $$

This matches Option 3: $ 1 - \alpha^2 - \beta\gamma = 0 $.

### Explanation of Incorrect Options:

- **Option 1: $ 1 + \alpha^2 + \beta\gamma = 0 $**

This would imply $ \alpha^2 + \beta\gamma = -1 $, which contradicts the requirement $ \alpha^2 + \beta\gamma = 1 $.

- **Option 2: $ 1 - \alpha^2 + \beta\gamma = 0 $**

This would imply $ \beta\gamma = \alpha^2 - 1 $, which does not satisfy the condition $ \alpha^2 + \beta\gamma = 1 $.

- **Option 4: $ 1 + \alpha^2 - \beta\gamma = 0 $**

This would imply $ \alpha^2 - \beta\gamma = -1 $, which again contradicts the requirement $ \alpha^2 + \beta\gamma = 1 $.

Thus, the correct answer is Option 3: $ 1 - \alpha^2 - \beta\gamma = 0 $.