`A=[(sinalpha,0),(0,sinalpha)] and det(A^2-1/2I)=0` , then a possible value of `alpha` is

Answer

To solve the problem, we need to find the possible value of \( \alpha \) given the matrix \( A \) and the condition \( \text{det}(A^2 - \frac{1}{2}I) = 0 \).

### Step 1: Define the matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} \sin \alpha & 0 \\ 0 & \sin \alpha \end{pmatrix} \]

### Step 2: Compute \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} \sin \alpha & 0 \\ 0 & \sin \alpha \end{pmatrix} \cdot \begin{pmatrix} \sin \alpha & 0 \\ 0 & \sin \alpha \end{pmatrix} \]
Calculating this gives:
\[ A^2 = \begin{pmatrix} \sin^2 \alpha & 0 \\ 0 & \sin^2 \alpha \end{pmatrix} \]

### Step 3: Define the identity matrix \( I \)
The identity matrix \( I \) in this case is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Thus, \( \frac{1}{2}I \) is:
\[ \frac{1}{2}I = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \]

### Step 4: Compute \( A^2 - \frac{1}{2}I \)
Now we compute \( A^2 - \frac{1}{2}I \):
\[ A^2 - \frac{1}{2}I = \begin{pmatrix} \sin^2 \alpha & 0 \\ 0 & \sin^2 \alpha \end{pmatrix} - \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \]
This results in:
\[ A^2 - \frac{1}{2}I = \begin{pmatrix} \sin^2 \alpha - \frac{1}{2} & 0 \\ 0 & \sin^2 \alpha - \frac{1}{2} \end{pmatrix} \]

### Step 5: Find the determinant
The determinant of a diagonal matrix is the product of its diagonal elements:
\[ \text{det}(A^2 - \frac{1}{2}I) = \left(\sin^2 \alpha - \frac{1}{2}\right) \cdot \left(\sin^2 \alpha - \frac{1}{2}\right) = \left(\sin^2 \alpha - \frac{1}{2}\right)^2 \]
Setting this equal to zero gives:
\[ \left(\sin^2 \alpha - \frac{1}{2}\right)^2 = 0 \]
This implies:
\[ \sin^2 \alpha - \frac{1}{2} = 0 \]
Thus:
\[ \sin^2 \alpha = \frac{1}{2} \]

### Step 6: Solve for \( \alpha \)
Taking the square root, we find:
\[ \sin \alpha = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]
The angles \( \alpha \) that satisfy this are:
\[ \alpha = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, a possible value of \( \alpha \) is:
\[ \alpha = \frac{\pi}{4} \]

### Final Answer
A possible value of \( \alpha \) is \( \frac{\pi}{4} \).