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, \( B \) is a skew-symmetric matrix, and \( A + B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \).

### Step 1: Define the properties of matrices A and B

Since \( A \) is symmetric, we have:
\[ A^T = A \]
And since \( B \) is skew-symmetric, we have:
\[ B^T = -B \]

### Step 2: Write the equation for \( A + B \)

We know:
\[ A + B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]

### Step 3: Take the transpose of both sides

Taking the transpose of both sides, we get:
\[ (A + B)^T = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix}^T \]
This simplifies to:
\[ A^T + B^T = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]
Substituting the properties of \( A \) and \( B \):
\[ A + (-B) = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]
This gives us:
\[ A - B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]

### Step 4: Set up the system of equations

Now we have two equations:
1. \( A + B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \) (Equation 1)
2. \( A - B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \) (Equation 2)

### Step 5: Add the two equations

Adding Equation 1 and Equation 2:
\[ (A + B) + (A - B) = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} + \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]
This simplifies to:
\[ 2A = \begin{pmatrix} 4 & 6 \\ 10 & -2 \end{pmatrix} \]

### Step 6: Solve for A

Dividing both sides by 2:
\[ A = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]

### Step 7: Substitute A back to find B

Now substitute \( A \) back into Equation 1:
\[ \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} + B = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \]
This implies:
\[ B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Step 8: Calculate the product AB

Now we can find \( AB \):
\[ AB = \begin{pmatrix} 2 & 3 \\ 5 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]
This results in:
\[ AB = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Final Answer

Thus, \( AB \) is equal to:
\[ \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]