If A and B are 2-rowed square matrics such that
`(A+B)=[(4,-3),(1,6)] and (A-B)=[(-2,-1),(5,2)]` then AB=?

Options

A. `[(-7,5),(1,-5)]`
B. `[(7,-5),(1,5)]`
C. `[(7,-1),(5,-5)]`
D. `[(7,-1),(-5,5)]`

Explanation

To solve the problem, we need to find the matrices A and B from the given equations and then compute the product AB.

Given:
1. \( A + B = \begin{pmatrix} 4 & -3 \\ 1 & 6 \end{pmatrix} \)
2. \( A - B = \begin{pmatrix} -2 & -1 \\ 5 & 2 \end{pmatrix} \)

### Step 1: Add the two equations
We can add the two equations to eliminate B:
\[ (A + B) + (A - B) = \begin{pmatrix} 4 & -3 \\ 1 & 6 \end{pmatrix} + \begin{pmatrix} -2 & -1 \\ 5 & 2 \end{pmatrix} \]
This simplifies to:
\[ 2A = \begin{pmatrix} 4 - 2 & -3 - 1 \\ 1 + 5 & 6 + 2 \end{pmatrix} = \begin{pmatrix} 2 & -4 \\ 6 & 8 \end{pmatrix} \]

### Step 2: Solve for A
Now, divide by 2 to find A:
\[ A = \frac{1}{2} \begin{pmatrix} 2 & -4 \\ 6 & 8 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} \]

### Step 3: Substitute A back to find B
Now, we can substitute A back into one of the original equations to find B. We'll use the first equation:
\[ A + B = \begin{pmatrix} 4 & -3 \\ 1 & 6 \end{pmatrix} \]
Substituting A:
\[ \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} + B = \begin{pmatrix} 4 & -3 \\ 1 & 6 \end{pmatrix} \]
Now, solve for B:
\[ B = \begin{pmatrix} 4 & -3 \\ 1 & 6 \end{pmatrix} - \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 4 - 1 & -3 + 2 \\ 1 - 3 & 6 - 4 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -2 & 2 \end{pmatrix} \]

### Step 4: Calculate the product AB
Now that we have both matrices A and B, we can calculate the product AB:
\[ AB = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -2 & 2 \end{pmatrix} \]
Calculating the product:
- First row, first column: \( 1 \cdot 3 + (-2) \cdot (-2) = 3 + 4 = 7 \)
- First row, second column: \( 1 \cdot (-1) + (-2) \cdot 2 = -1 - 4 = -5 \)
- Second row, first column: \( 3 \cdot 3 + 4 \cdot (-2) = 9 - 8 = 1 \)
- Second row, second column: \( 3 \cdot (-1) + 4 \cdot 2 = -3 + 8 = 5 \)

Thus,
\[ AB = \begin{pmatrix} 7 & -5 \\ 1 & 5 \end{pmatrix} \]

### Final Answer
The product \( AB \) is:
\[ \begin{pmatrix} 7 & -5 \\ 1 & 5 \end{pmatrix} \]