Let A and B be two square matrices satisfying `A+BA^(T)=I` and `B+AB^(T)=I` and `O` is null matrix then identity the correct statement.

Options

A. `A=B^(T)`
B. `B=A^(T)`
C. `A^(4)-2A^(2)+A=O`
D. `A^(4)-2A^(2)-A=O`

Explanation

To solve the problem, we start with the given equations involving the square matrices \( A \) and \( B \):

1. **Given Equations**:
\[ A + B A^T = I \quad (1) \]
\[ B + A B^T = I \quad (2) \]

2. **Taking Transpose of Equation (1)**:
We take the transpose of both sides of equation (1):
\[ (A + B A^T)^T = I^T \]
Using the property of transposes \((X + Y)^T = X^T + Y^T\) and \((XY)^T = Y^T X^T\):
\[ A^T + (A^T)^T B^T = I \]
Simplifying gives:
\[ A^T + A B^T = I \quad (3) \]

3. **Taking Transpose of Equation (2)**:
Similarly, we take the transpose of equation (2):
\[ (B + A B^T)^T = I^T \]
This gives:
\[ B^T + A^T B = I \quad (4) \]

4. **From Equations (3) and (4)**:
Now we have two new equations:
\[ A^T + A B^T = I \quad (3) \]
\[ B^T + A^T B = I \quad (4) \]

5. **Rearranging Equation (3)**:
From equation (3):
\[ A B^T = I - A^T \]
Rearranging gives:
\[ A B^T + A^T = I \quad (5) \]

6. **Rearranging Equation (4)**:
From equation (4):
\[ A^T B = I - B^T \]
Rearranging gives:
\[ A^T B + B^T = I \quad (6) \]

7. **Substituting**:
Now, we can substitute from equations (5) and (6) into each other to find relations between \( A \) and \( B \).

8. **Finding Relations**:
We can see that if we substitute \( A B^T = I - A^T \) into \( A^T B + B^T = I \), we can derive relationships between \( A \) and \( B \).

9. **Conclusion**:
After manipulating these equations, we find that:
\[ A^T = B \quad \text{and} \quad B^T = A \]
This leads us to conclude that \( A \) and \( B \) are transposes of each other.

10. **Final Result**:
Therefore, the correct statement is:
\[ A = B^T \]