Let A denote the matrix `({:(0,i),(i,0):})`, where ` i^(2) = -1` , and let `I` denote the identity matrix `({:(1,0),(0,1):})`. Then `I + A + A^(2) + "….." + A^(2010)` is -

Options

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

Explanation

To solve the problem, we need to evaluate the expression \( I + A + A^2 + \ldots + A^{2010} \) where \( A = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \) and \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

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

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

### 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
We observe that:
- \( A^0 = I \)
- \( A^1 = A \)
- \( A^2 = -I \)
- \( A^3 = -A \)
- \( A^4 = I \)

This pattern repeats every 4 terms:
- \( A^{4k} = I \)
- \( A^{4k+1} = A \)
- \( A^{4k+2} = -I \)
- \( A^{4k+3} = -A \)

### Step 5: Sum the series
We need to sum \( I + A + A^2 + A^3 + A^4 + \ldots + A^{2010} \).

Since the powers of \( A \) repeat every 4 terms, we can group the terms:
\[ (I + A + (-I) + (-A)) = 0 \]
This means every complete set of 4 terms sums to 0.

### Step 6: Count the complete sets
To find how many complete sets of 4 fit into 2011 terms (from \( A^0 \) to \( A^{2010} \)):
\[ \text{Number of complete sets} = \left\lfloor \frac{2011}{4} \right\rfloor = 502 \]
This accounts for \( 502 \times 4 = 2008 \) terms.

### Step 7: Remaining terms
The remaining terms are:
- \( A^{2009} = A \)
- \( A^{2010} = A^2 = -I \)

### Step 8: Final sum
Thus, the total sum becomes:
\[ \text{Total sum} = 502 \cdot 0 + A + (-I) = A - I \]
Substituting the values:
\[ A - I = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & i \\ i & -1 \end{pmatrix} \]

### Final Answer
The final result is:
\[ \begin{pmatrix} -1 & i \\ i & -1 \end{pmatrix} \]