If matrix `A=[(0,-1),(1,0)]`, then `A^16` =

Options

A. `[(0,-1),(1,0)]`
B. `[(0,1),(1,0)]`
C. `[(-1,0),(0,1)]`
D. `[(1,0),(0,1)]`

Answer

To find \( A^{16} \) where \( A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \), we can follow these steps:

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

We start by calculating \( A^2 = A \cdot A \).

\[ A^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]

Calculating the product:

- First row, first column: \( 0 \cdot 0 + (-1) \cdot 1 = -1 \)
- First row, second column: \( 0 \cdot (-1) + (-1) \cdot 0 = 0 \)
- Second row, first column: \( 1 \cdot 0 + 0 \cdot 1 = 0 \)
- Second row, second column: \( 1 \cdot (-1) + 0 \cdot 0 = -1 \)

Thus,

\[ A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I \]

### Step 2: Calculate \( A^4 \)

Next, we calculate \( A^4 \) using \( A^2 \):

\[ A^4 = A^2 \cdot A^2 = (-I) \cdot (-I) = I \]

### Step 3: Calculate \( A^{16} \)

Now, we can find \( A^{16} \):

\[ A^{16} = (A^4)^4 = I^4 = I \]

### Conclusion

Thus,

\[ A^{16} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Final Answer

The answer is:

\[ A^{16} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]