Let `A`, `B` are square matrices of same order satisfying `AB=A` and `BA=B` then `(A^(2010)+B^(2010))^(2011)` equals.

Options

A. `A+B`
B. `2010(A+B)`
C. `2011(A+B)`
D. `2^(2011)(A+B)`

Explanation

To solve the problem, we need to analyze the given equations involving the matrices \( A \) and \( B \):

1. **Given Equations**:
- \( AB = A \)
- \( BA = B \)

2. **Understanding the Implications**:
- From \( AB = A \), we can rearrange this to \( AB - A = 0 \) or \( A(B - I) = 0 \).
- From \( BA = B \), we can rearrange this to \( BA - B = 0 \) or \( (B - I)A = 0 \).

This suggests that \( A \) and \( B \) might be projections. Specifically, if we assume \( A \) and \( B \) are idempotent matrices (i.e., \( A^2 = A \) and \( B^2 = B \)), we can derive further properties.

3. **Finding \( A^n \) and \( B^n \)**:
- If \( A^2 = A \), then for any positive integer \( n \), we have \( A^n = A \).
- Similarly, if \( B^2 = B \), then \( B^n = B \).

Therefore, we can conclude:
\[ A^{2010} = A \quad \text{and} \quad B^{2010} = B \]

4. **Substituting Back**:
- Now, substituting these results into the expression:
\[ A^{2010} + B^{2010} = A + B \]

5. **Final Expression**:
- We need to evaluate \( (A + B)^{2011} \).
- Since \( A + B \) is a linear combination of the matrices \( A \) and \( B \), we can raise it to the power of 2011.

6. **Conclusion**:
- Thus, we have:
\[ (A^{2010} + B^{2010})^{2011} = (A + B)^{2011} \]

### Final Answer:
\[ (A^{2010} + B^{2010})^{2011} = (A + B)^{2011} \]