Let `A = [{:(5,6,1),(2,-1,5):}]`. Let there exist a matrix B such that `AB = [{:(35,49),(29,13):}]`. What is B equal to ?

Options

A. `[{:(5,1,4),(2,6,3):}]`
B. `[{:(2,6,3),(5,1,4):}]`
C. `[{:(5,2),(1,6),(4,3):}]`
D. `[{:(2,5),(6,1),(3,4):}]`

Explanation

To find the matrix \( B \) such that \( AB = \begin{pmatrix} 35 & 49 \\ 29 & 13 \end{pmatrix} \), we will follow these steps:

### Step 1: Define the matrices
Given:
\[ A = \begin{pmatrix} 5 & 6 & 1 \\ 2 & -1 & 5 \end{pmatrix} \]
Let \( B \) be represented as:
\[ B = \begin{pmatrix} a & b \\ c & d \\ e & f \end{pmatrix} \]
where \( a, b, c, d, e, f \) are the unknowns we need to determine.

### Step 2: Determine the dimensions of \( B \)
Since \( A \) is a \( 2 \times 3 \) matrix and the resulting matrix \( AB \) is a \( 2 \times 2 \) matrix, \( B \) must be a \( 3 \times 2 \) matrix.

### Step 3: Perform matrix multiplication
The multiplication \( AB \) is calculated as follows:
\[ AB = \begin{pmatrix} 5 & 6 & 1 \\ 2 & -1 & 5 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \\ e & f \end{pmatrix} \]
Calculating the elements of the resulting matrix:
1. First row, first column:
\[ 5a + 6c + 1e \]
2. First row, second column:
\[ 5b + 6d + 1f \]
3. Second row, first column:
\[ 2a - 1c + 5e \]
4. Second row, second column:
\[ 2b - 1d + 5f \]

Thus, we have:
\[ AB = \begin{pmatrix} 5a + 6c + e & 5b + 6d + f \\ 2a - c + 5e & 2b - d + 5f \end{pmatrix} \]

### Step 4: Set up equations
We know that:
\[ AB = \begin{pmatrix} 35 & 49 \\ 29 & 13 \end{pmatrix} \]
This gives us the following equations:
1. \( 5a + 6c + e = 35 \) (Equation 1)
2. \( 5b + 6d + f = 49 \) (Equation 2)
3. \( 2a - c + 5e = 29 \) (Equation 3)
4. \( 2b - d + 5f = 13 \) (Equation 4)

### Step 5: Solve the equations
We will solve these equations step by step.

**From Equation 1:**
\[ e = 35 - 5a - 6c \]

**Substituting \( e \) into Equation 3:**
\[ 2a - c + 5(35 - 5a - 6c) = 29 \]
Expanding and simplifying:
\[ 2a - c + 175 - 25a - 30c = 29 \]
\[ -23a - 31c + 175 = 29 \]
\[ -23a - 31c = -146 \quad \text{(Equation 5)} \]

**From Equation 2:**
\[ f = 49 - 5b - 6d \]

**Substituting \( f \) into Equation 4:**
\[ 2b - d + 5(49 - 5b - 6d) = 13 \]
Expanding and simplifying:
\[ 2b - d + 245 - 25b - 30d = 13 \]
\[ -23b - 31d + 245 = 13 \]
\[ -23b - 31d = -232 \quad \text{(Equation 6)} \]

### Step 6: Solve Equations 5 and 6
We can solve Equations 5 and 6 simultaneously.

**From Equation 5:**
\[ 23a + 31c = 146 \]

**From Equation 6:**
\[ 23b + 31d = 232 \]

Now, we can choose values for \( a, b, c, d \) that satisfy these equations. After testing some integer values, we find:
- \( a = 5 \)
- \( b = 2 \)
- \( c = 1 \)
- \( d = 6 \)
- \( e = 4 \)
- \( f = 3 \)

### Step 7: Write the final matrix \( B \)
Thus, the matrix \( B \) is:
\[ B = \begin{pmatrix} 5 & 2 \\ 1 & 6 \\ 4 & 3 \end{pmatrix} \]