If `A=[a_(ij)]_(4xx3)`, where `a_(ij)=(i-j)/(i+j)`, then find A.

Options

A. `[(0,-1//3,-1//2),(1//2,0,1//5),(1//3,1//5,0),(3//5,1//3,1//7)]`
B. `[(0,-1//3,-1//2),(1//3,0,-1//5),(1//2,1//5,0),(3//5,1//3,1//7)]`
C. `[(0,-3,-1//2),(2,0,5),(3,5,0),(3//5,3,7)]`
D. `[(0,1//3,1//2),(-1//3,0,1//5),(-1//2,-1//5,0),(-3//5,-1//3,-1//7)]`

Answer

To find the matrix \( A = [a_{ij}]_{4 \times 3} \) where \( a_{ij} = \frac{i - j}{i + j} \), we will calculate the elements of the matrix for \( i = 1, 2, 3, 4 \) and \( j = 1, 2, 3 \).

### Step 1: Define the matrix dimensions
The matrix \( A \) has 4 rows and 3 columns, so we will calculate \( a_{ij} \) for \( i = 1, 2, 3, 4 \) and \( j = 1, 2, 3 \).

### Step 2: Calculate the elements of the matrix
Now we will compute each element of the matrix \( A \):

- For \( i = 1 \):
- \( a_{11} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \)
- \( a_{12} = \frac{1 - 2}{1 + 2} = \frac{-1}{3} \)
- \( a_{13} = \frac{1 - 3}{1 + 3} = \frac{-2}{4} = -\frac{1}{2} \)

- For \( i = 2 \):
- \( a_{21} = \frac{2 - 1}{2 + 1} = \frac{1}{3} \)
- \( a_{22} = \frac{2 - 2}{2 + 2} = \frac{0}{4} = 0 \)
- \( a_{23} = \frac{2 - 3}{2 + 3} = \frac{-1}{5} \)

- For \( i = 3 \):
- \( a_{31} = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2} \)
- \( a_{32} = \frac{3 - 2}{3 + 2} = \frac{1}{5} \)
- \( a_{33} = \frac{3 - 3}{3 + 3} = \frac{0}{6} = 0 \)

- For \( i = 4 \):
- \( a_{41} = \frac{4 - 1}{4 + 1} = \frac{3}{5} \)
- \( a_{42} = \frac{4 - 2}{4 + 2} = \frac{2}{6} = \frac{1}{3} \)
- \( a_{43} = \frac{4 - 3}{4 + 3} = \frac{1}{7} \)

### Step 3: Assemble the matrix
Now we can assemble the matrix \( A \) using the calculated elements:

\[ A = \begin{bmatrix} 0 & -\frac{1}{3} & -\frac{1}{2} \\ \frac{1}{3} & 0 & -\frac{1}{5} \\ \frac{1}{2} & \frac{1}{5} & 0 \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{bmatrix} \]

### Final Result
Thus, the matrix \( A \) is:

\[ A = \begin{bmatrix} 0 & -\frac{1}{3} & -\frac{1}{2} \\ \frac{1}{3} & 0 & -\frac{1}{5} \\ \frac{1}{2} & \frac{1}{5} & 0 \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{bmatrix} \]