If A and Bare square matrices of same order such that AB = A and BA = B, then `AB^(2)+B^(2)=`:

Options

A. AB
B. A+B
C. 2AB
D. 2BA

Answer

To solve the problem, we are given two square matrices \( A \) and \( B \) of the same order such that \( AB = A \) and \( BA = B \). We need to find the expression \( AB^2 + B^2 \).

### Step-by-Step Solution:

1. **Start with the given expression**:
\[ AB^2 + B^2 \]

2. **Rewrite \( B^2 \)**:
We can express \( B^2 \) as \( B \cdot B \). Thus, we rewrite the expression:
\[ AB^2 + B^2 = AB \cdot B + B \cdot B \]

3. **Use the property \( AB = A \)**:
From the given condition \( AB = A \), we can substitute \( A \) for \( AB \):
\[ A \cdot B + B^2 \]

4. **Now, express \( B^2 \)**:
To find \( B^2 \), we can use the second condition \( BA = B \). Multiply both sides of \( BA = B \) by \( B \):
\[ B \cdot A \cdot B = B \cdot B \]
This gives us:
\[ B^2 = B \]

5. **Substituting \( B^2 \) back into the expression**:
Now we substitute \( B^2 \) with \( B \):
\[ A + B \]

6. **Final Result**:
Therefore, the expression \( AB^2 + B^2 \) simplifies to:
\[ A + B \]

### Conclusion:
The final answer is:
\[ AB^2 + B^2 = A + B \]