If A is a symmetric matrix and B is a skew-symmetric matrix such that
`A + B = [{:(2,3),(5,-1):}]`, then AB is equal to

Options

A. `[(-4, 2),(1,4)]`
B. `[(4,-2),(-1,-4)]`
C. `[(4, -2),(1,-4)]`
D. `[(-4, -2),(-1, 4)]`

Explanation

To solve the problem, we need to find the product \( AB \) given that \( A \) is a symmetric matrix and \( B \) is a skew-symmetric matrix such that \( A + B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \).

### Step 1: Define the matrices
Let \( P = A + B \). We know that:
\[ P = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]

### Step 2: Express \( A \) and \( B \)
Since \( A \) is symmetric and \( B \) is skew-symmetric, we can express them as:
\[ A = \frac{1}{2}(P + P^T) \]
\[ B = \frac{1}{2}(P - P^T) \]

### Step 3: Calculate \( P^T \)
First, we need to find the transpose of \( P \):
\[ P^T = \begin{pmatrix} 2 & 5 \\ 3 & -1 \end{pmatrix} \]

### Step 4: Calculate \( A \)
Now, we can find \( A \):
\[ A = \frac{1}{2} \left( \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} + \begin{pmatrix} 2 & 5 \\ 3 & -1 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 4 & 8 \\ 8 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 4 & -1 \end{pmatrix} \]

### Step 5: Calculate \( B \)
Next, we find \( B \):
\[ B = \frac{1}{2} \left( \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} - \begin{pmatrix} 2 & 5 \\ 3 & -1 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]

### Step 6: Calculate \( AB \)
Now we can find the product \( AB \):
\[ AB = \begin{pmatrix} 2 & 4 \\ 4 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]
Calculating the product:
\[ AB = \begin{pmatrix} 2 \cdot 0 + 4 \cdot 1 & 2 \cdot -1 + 4 \cdot 0 \\ 4 \cdot 0 + (-1) \cdot 1 & 4 \cdot -1 + (-1) \cdot 0 \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ -1 & -4 \end{pmatrix} \]

### Final Result
Thus, the product \( AB \) is:
\[ AB = \begin{pmatrix} 4 & -2 \\ -1 & -4 \end{pmatrix} \]