If `A = [(cos alpha,sin alpha),(-sin alpha,cos alpha)]`, then `A^(10)` = .........

Options

A. `[(cos 10 alpha,-sin 10 alpha),(sin 10 alpha,cos 10 alpha)]`
B. `[(cos 10 alpha,sin 10 alpha),(-sin 10 alpha,cos 10 alpha)]`
C. `[(cos 10 alpha,sin 10 alpha),(-sin 10 alpha,-cos 10 alpha)]`
D. `[(cos 10 alpha,-sin 10 alpha),(-sin 10 alpha,-cos 10 alpha)]`

Answer

To find \( A^{10} \) for the matrix

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

we will first compute \( A^2 \) and then generalize the result to \( A^{10} \).

### Step 1: Compute \( A^2 \)

To compute \( A^2 \), we multiply \( A \) by itself:

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

Calculating the elements of \( A^2 \):

- The element at (1,1):
\[ \cos \alpha \cdot \cos \alpha + \sin \alpha \cdot (-\sin \alpha) = \cos^2 \alpha - \sin^2 \alpha. \]

- The element at (1,2):
\[ \cos \alpha \cdot \sin \alpha + \sin \alpha \cdot \cos \alpha = 2 \cos \alpha \sin \alpha. \]

- The element at (2,1):
\[ -\sin \alpha \cdot \cos \alpha + \cos \alpha \cdot (-\sin \alpha) = -2 \sin \alpha \cos \alpha. \]

- The element at (2,2):
\[ -\sin \alpha \cdot \sin \alpha + \cos \alpha \cdot \cos \alpha = \cos^2 \alpha - \sin^2 \alpha. \]

Thus, we have:

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

Using the double angle identities, we can rewrite this as:

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

### Step 2: Generalize to \( A^n \)

From the pattern observed, we can generalize:

\[ A^n = \begin{pmatrix} \cos(n\alpha) & \sin(n\alpha) \\ -\sin(n\alpha) & \cos(n\alpha) \end{pmatrix}. \]

### Step 3: Compute \( A^{10} \)

Now, substituting \( n = 10 \):

\[ A^{10} = \begin{pmatrix} \cos(10\alpha) & \sin(10\alpha) \\ -\sin(10\alpha) & \cos(10\alpha) \end{pmatrix}. \]

### Final Answer

Thus, the final result is:

\[ A^{10} = \begin{pmatrix} \cos(10\alpha) & \sin(10\alpha) \\ -\sin(10\alpha) & \cos(10\alpha) \end{pmatrix}. \]

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