`A=[{:(2,4),(3,2):}],B=[{:(1,3),(-2,5):}],C=[{:(-2,5),(3,4):}]`
find each of the following :
(i) A+B (ii) A-B
(iii) 3A-C (iv) AB
(V) BA

Explanation

To solve the problem, we will perform the following calculations step by step:

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

### (i) Finding \( A + B \)

To add two matrices, we add their corresponding elements:

\[ A + B = \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} 2 + 1 & 4 + 3 \\ 3 - 2 & 2 + 5 \end{pmatrix} \]

Calculating the elements:

\[ = \begin{pmatrix} 3 & 7 \\ 1 & 7 \end{pmatrix} \]

### (ii) Finding \( A - B \)

To subtract two matrices, we subtract their corresponding elements:

\[ A - B = \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} 2 - 1 & 4 - 3 \\ 3 + 2 & 2 - 5 \end{pmatrix} \]

Calculating the elements:

\[ = \begin{pmatrix} 1 & 1 \\ 5 & -3 \end{pmatrix} \]

### (iii) Finding \( 3A - C \)

First, we multiply matrix \( A \) by 3:

\[ 3A = 3 \times \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 6 & 12 \\ 9 & 6 \end{pmatrix} \]

Now, we subtract matrix \( C \):

\[ 3A - C = \begin{pmatrix} 6 & 12 \\ 9 & 6 \end{pmatrix} - \begin{pmatrix} -2 & 5 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 6 + 2 & 12 - 5 \\ 9 - 3 & 6 - 4 \end{pmatrix} \]

Calculating the elements:

\[ = \begin{pmatrix} 8 & 7 \\ 6 & 2 \end{pmatrix} \]

### (iv) Finding \( AB \)

To multiply matrices \( A \) and \( B \):

\[ AB = \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix} \times \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 2 \times 1 + 4 \times (-2) = 2 - 8 = -6 \)
- First row, second column: \( 2 \times 3 + 4 \times 5 = 6 + 20 = 26 \)
- Second row, first column: \( 3 \times 1 + 2 \times (-2) = 3 - 4 = -1 \)
- Second row, second column: \( 3 \times 3 + 2 \times 5 = 9 + 10 = 19 \)

Thus,

\[ AB = \begin{pmatrix} -6 & 26 \\ -1 & 19 \end{pmatrix} \]

### (v) Finding \( BA \)

To multiply matrices \( B \) and \( A \):

\[ BA = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} \times \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 1 \times 2 + 3 \times 3 = 2 + 9 = 11 \)
- First row, second column: \( 1 \times 4 + 3 \times 2 = 4 + 6 = 10 \)
- Second row, first column: \( -2 \times 2 + 5 \times 3 = -4 + 15 = 11 \)
- Second row, second column: \( -2 \times 4 + 5 \times 2 = -8 + 10 = 2 \)

Thus,

\[ BA = \begin{pmatrix} 11 & 10 \\ 11 & 2 \end{pmatrix} \]

### Summary of Results

1. \( A + B = \begin{pmatrix} 3 & 7 \\ 1 & 7 \end{pmatrix} \)
2. \( A - B = \begin{pmatrix} 1 & 1 \\ 5 & -3 \end{pmatrix} \)
3. \( 3A - C = \begin{pmatrix} 8 & 7 \\ 6 & 2 \end{pmatrix} \)
4. \( AB = \begin{pmatrix} -6 & 26 \\ -1 & 19 \end{pmatrix} \)
5. \( BA = \begin{pmatrix} 11 & 10 \\ 11 & 2 \end{pmatrix} \)