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 \( A^{16} \) for the matrix

\[ A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \]

we can follow these steps:

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

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

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

Calculating the product:

\[ = \begin{pmatrix} (0 \cdot 0 + -1 \cdot 1) & (0 \cdot -1 + -1 \cdot 0) \\ (1 \cdot 0 + 0 \cdot 1) & (1 \cdot -1 + 0 \cdot 0) \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

Thus,

\[ A^2 = -I \]

where \( I \) is the identity matrix:

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

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

Now, we can find \( A^4 \):

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

Since \( (-I)^2 = I \):

\[ A^4 = I \]

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

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

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

Since the identity matrix raised to any power is still the identity matrix:

\[ I^4 = I \]

### Final Result

Thus,

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

### Summary

The final answer is:

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

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