if A and B are symmetric matrices of the same order and `P=AB+BA and Q=AB-BA,` then `(PQ)'` is equal to

Options

A. PQ
B. QP
C. `-QP`
D. none of these

Answer

To solve the problem, we need to find the transpose of the product \( PQ \) where \( P = AB + BA \) and \( Q = AB - BA \) for symmetric matrices \( A \) and \( B \).

### Step-by-Step Solution:

1. **Understanding the Properties of Symmetric Matrices**:
Since \( A \) and \( B \) are symmetric matrices, we have:
\[ A^T = A \quad \text{and} \quad B^T = B \]

2. **Define \( P \) and \( Q \)**:
We are given:
\[ P = AB + BA \]
\[ Q = AB - BA \]

3. **Finding \( P^T \)**:
To find \( P^T \):
\[ P^T = (AB + BA)^T = (AB)^T + (BA)^T \]
Using the property of transposes, we have:
\[ (AB)^T = B^T A^T = BA \quad \text{and} \quad (BA)^T = A^T B^T = AB \]
Therefore:
\[ P^T = BA + AB = AB + BA = P \]
Thus, we conclude:
\[ P^T = P \]

4. **Finding \( Q^T \)**:
Now, we find \( Q^T \):
\[ Q^T = (AB - BA)^T = (AB)^T - (BA)^T \]
Again, using the transpose properties:
\[ Q^T = BA - AB \]
This can be rewritten as:
\[ Q^T = - (AB - BA) = -Q \]
Thus, we conclude:
\[ Q^T = -Q \]

5. **Finding \( (PQ)^T \)**:
Now, we need to find \( (PQ)^T \):
\[ (PQ)^T = Q^T P^T \]
Substituting the results from steps 3 and 4:
\[ (PQ)^T = (-Q) P = -QP \]

6. **Final Result**:
Therefore, we have:
\[ (PQ)^T = -QP \]

### Conclusion:
The answer is that \( (PQ)^T \) is equal to \( -QP \).