Let `A=[(0, i),(i, 0)]`, where `i^(2)=-1`. Let I denotes the identity matrix of order 2, then `I+A+A^(2)+A^(3)+……..A^(110)` is equal to

Options

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

Answer

To solve the problem, we need to evaluate the expression \( I + A + A^2 + A^3 + \ldots + A^{110} \), where \( A = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \) and \( I \) is the identity matrix of order 2.

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

We start by calculating \( A^2 \):

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

Calculating the elements:

- First row, first column: \( 0 \cdot 0 + i \cdot i = i^2 = -1 \)
- First row, second column: \( 0 \cdot i + i \cdot 0 = 0 \)
- Second row, first column: \( i \cdot 0 + 0 \cdot i = 0 \)
- Second row, second column: \( i \cdot i + 0 \cdot 0 = i^2 = -1 \)

Thus, we have:

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

### Step 2: Calculate \( A^3 \)

Next, we calculate \( A^3 \):

\[ A^3 = A^2 \cdot A = (-I) \cdot A = -A = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \]

### Step 3: Calculate \( A^4 \)

Now, we calculate \( A^4 \):

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

### Step 4: Identify the pattern

From our calculations, we observe the following pattern:

- \( A^0 = I \)
- \( A^1 = A \)
- \( A^2 = -I \)
- \( A^3 = -A \)
- \( A^4 = I \)
- \( A^5 = A \)
- \( A^6 = -I \)
- \( A^7 = -A \)
- ...

This pattern repeats every 4 terms. Thus, we can summarize:

- \( A^{4n} = I \)
- \( A^{4n+1} = A \)
- \( A^{4n+2} = -I \)
- \( A^{4n+3} = -A \)

### Step 5: Calculate \( I + A + A^2 + A^3 + \ldots + A^{110} \)

Now, we need to find how many complete cycles of 4 fit into 110:

\[ 110 = 4 \times 27 + 2 \]

This means we have 27 complete cycles of \( I, A, -I, -A \) and then the additional terms \( I + A \) from the next cycle.

Calculating the contribution from the complete cycles:

Each complete cycle contributes:

\[ I + A - I - A = 0 \]

Thus, the contribution from 27 cycles is:

\[ 27 \times 0 = 0 \]

Now, we add the remaining terms:

\[ I + A \]

### Final Result

Therefore, the final result is:

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

Thus, the answer is:

\[ \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \]