`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

Let's solve the given matrices step by step.

Given matrices:
\[ A = \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}, \quad C = \begin{pmatrix} -2 & 5 \\ 3 & 4 \end{pmatrix} \]

### (i) A + B
To find \( A + B \), we add the corresponding elements of matrices A and B.

\[ A + B = \begin{pmatrix} 2 + 1 & 4 + 3 \\ 3 - 2 & 2 + 5 \end{pmatrix} = \begin{pmatrix} 3 & 7 \\ 1 & 7 \end{pmatrix} \]

### (ii) A - B
To find \( A - B \), we subtract the corresponding elements of matrix B from matrix A.

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

### (iii) 3A - C
First, we calculate \( 3A \) by multiplying each element of matrix A by 3.

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

Now, we subtract matrix C from \( 3A \).

\[ 3A - C = \begin{pmatrix} 6 - (-2) & 12 - 5 \\ 9 - 3 & 6 - 4 \end{pmatrix} = \begin{pmatrix} 8 & 7 \\ 6 & 2 \end{pmatrix} \]

### (iv) AB
To find \( AB \), we perform matrix multiplication.

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

Calculating the elements:

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

So,

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

### (v) BA
Now, we find \( BA \) by multiplying matrices B and A.

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

Calculating the elements:

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

So,

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

### 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. \( BA = \begin{pmatrix} 11 & 10 \\ 11 & 2 \end{pmatrix} \)