If `A_(alpha)=[(cosalpha,sinalpha),(-sinalpha,cosalpha)] then (A_(alpha))^2=?`

Options

A. `[(cos^2alpha,sin^2alpha),(-sin^2alpha,cos^2alpha)]`
B. `[(cos2alpha,sin2alpha),(-sin2alpha,cos2alpha)]`
C. `[(2cosalpha,2sinalpha),(-sinalpha,2cosalpha)]`
D. none of these

Answer

To solve the problem, we need to find the square of the matrix \( A_{\alpha} \) given by:

\[ A_{\alpha} = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \]

### Step 1: Write down the expression for \( A_{\alpha}^2 \)

We need to compute \( A_{\alpha}^2 = A_{\alpha} \cdot A_{\alpha} \).

\[ A_{\alpha}^2 = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \cdot \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \]

### Step 2: Perform matrix multiplication

Using the formula for matrix multiplication, we calculate each element of the resulting matrix:

1. **First row, first column:**
\[ \cos \alpha \cdot \cos \alpha + \sin \alpha \cdot (-\sin \alpha) = \cos^2 \alpha - \sin^2 \alpha \]

2. **First row, second column:**
\[ \cos \alpha \cdot \sin \alpha + \sin \alpha \cdot \cos \alpha = \cos \alpha \sin \alpha + \sin \alpha \cos \alpha = 2 \sin \alpha \cos \alpha \]

3. **Second row, first column:**
\[ -\sin \alpha \cdot \cos \alpha + \cos \alpha \cdot (-\sin \alpha) = -\sin \alpha \cos \alpha - \sin \alpha \cos \alpha = -2 \sin \alpha \cos \alpha \]

4. **Second row, second column:**
\[ -\sin \alpha \cdot \sin \alpha + \cos \alpha \cdot \cos \alpha = -\sin^2 \alpha + \cos^2 \alpha = \cos^2 \alpha - \sin^2 \alpha \]

### Step 3: Combine the results

Putting all these results together, we have:

\[ A_{\alpha}^2 = \begin{pmatrix} \cos^2 \alpha - \sin^2 \alpha & 2 \sin \alpha \cos \alpha \\ -2 \sin \alpha \cos \alpha & \cos^2 \alpha - \sin^2 \alpha \end{pmatrix} \]

### Step 4: Use trigonometric identities

We can simplify the matrix using the double angle formulas:

- \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \)
- \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \)

Thus, we can rewrite \( A_{\alpha}^2 \) as:

\[ A_{\alpha}^2 = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{pmatrix} \]

### Final Result

Therefore, the final answer is:

\[ A_{\alpha}^2 = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{pmatrix} \]

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