If `{:A=[(i,0),(0,i)]:},n in N" then " A^(4n)`equal

Options

A. `{:[(0,i),(i,0)]:}`
B. `{:[(0,0),(0,0)]:}`
C. `{:[(1,0),(0,1)]:}`
D. `{:[(0,i),(i,0)]:}`

Explanation

To solve the problem, we need to find \( A^{4n} \) where \( A = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix} \) and \( n \) is a natural number.

### Step-by-Step Solution:

**Step 1: Find \( A^2 \)**
We start by calculating \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix} \cdot \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix} \]
Calculating the product:
\[ = \begin{pmatrix} i \cdot i & 0 \\ 0 & i \cdot i \end{pmatrix} = \begin{pmatrix} i^2 & 0 \\ 0 & i^2 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I \]
where \( I \) is the identity matrix.

**Step 2: Find \( A^4 \)**
Next, we calculate \( A^4 \):
\[ A^4 = A^2 \cdot A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]
Calculating the product:
\[ = \begin{pmatrix} (-1)(-1) & 0 \\ 0 & (-1)(-1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

**Step 3: Find \( A^{4n} \)**
Now, we can find \( A^{4n} \):
\[ A^{4n} = (A^4)^n = I^n \]
Since the identity matrix raised to any power is still the identity matrix:
\[ I^n = I \]

### Final Answer:
Thus, \( A^{4n} = I \), where \( I \) is the identity matrix.

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