`if A=[{:(2,8),(-7,3):}]and B=[{:(-1,3),(2,-4):}]`, then show that : (i) (A+B)'=A+B' (ii) (A+2B)'=A'+2B' (iii) (AB)'=B'A'

Explanation

To solve the given problem, we will verify three statements involving the matrices \( A \) and \( B \).

Given:
\[ A = \begin{pmatrix} 2 & 8 \\ -7 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 3 \\ 2 & -4 \end{pmatrix} \]

### (i) Show that \( (A + B)' = A + B' \)

**Step 1: Calculate \( A + B \)**

\[ A + B = \begin{pmatrix} 2 & 8 \\ -7 & 3 \end{pmatrix} + \begin{pmatrix} -1 & 3 \\ 2 & -4 \end{pmatrix} = \begin{pmatrix} 2 + (-1) & 8 + 3 \\ -7 + 2 & 3 + (-4) \end{pmatrix} = \begin{pmatrix} 1 & 11 \\ -5 & -1 \end{pmatrix} \]

**Step 2: Calculate \( (A + B)' \)**

\[ (A + B)' = \begin{pmatrix} 1 & 11 \\ -5 & -1 \end{pmatrix}' = \begin{pmatrix} 1 & -5 \\ 11 & -1 \end{pmatrix} \]

**Step 3: Calculate \( B' \)**

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

**Step 4: Calculate \( A + B' \)**

\[ A + B' = \begin{pmatrix} 2 & 8 \\ -7 & 3 \end{pmatrix} + \begin{pmatrix} -1 & 2 \\ 3 & -4 \end{pmatrix} = \begin{pmatrix} 2 + (-1) & 8 + 2 \\ -7 + 3 & 3 + (-4) \end{pmatrix} = \begin{pmatrix} 1 & 10 \\ -4 & -1 \end{pmatrix} \]

**Step 5: Compare \( (A + B)' \) and \( A + B' \)**

\[ (A + B)' = \begin{pmatrix} 1 & -5 \\ 11 & -1 \end{pmatrix} \quad \text{and} \quad A + B' = \begin{pmatrix} 1 & 10 \\ -4 & -1 \end{pmatrix} \]

Since \( (A + B)' \neq A + B' \), the first statement is incorrect.

### (ii) Show that \( (A + 2B)' = A' + 2B' \)

**Step 1: Calculate \( 2B \)**

\[ 2B = 2 \cdot \begin{pmatrix} -1 & 3 \\ 2 & -4 \end{pmatrix} = \begin{pmatrix} -2 & 6 \\ 4 & -8 \end{pmatrix} \]

**Step 2: Calculate \( A + 2B \)**

\[ A + 2B = \begin{pmatrix} 2 & 8 \\ -7 & 3 \end{pmatrix} + \begin{pmatrix} -2 & 6 \\ 4 & -8 \end{pmatrix} = \begin{pmatrix} 2 + (-2) & 8 + 6 \\ -7 + 4 & 3 + (-8) \end{pmatrix} = \begin{pmatrix} 0 & 14 \\ -3 & -5 \end{pmatrix} \]

**Step 3: Calculate \( (A + 2B)' \)**

\[ (A + 2B)' = \begin{pmatrix} 0 & 14 \\ -3 & -5 \end{pmatrix}' = \begin{pmatrix} 0 & -3 \\ 14 & -5 \end{pmatrix} \]

**Step 4: Calculate \( A' \)**

\[ A' = \begin{pmatrix} 2 & 8 \\ -7 & 3 \end{pmatrix}' = \begin{pmatrix} 2 & -7 \\ 8 & 3 \end{pmatrix} \]

**Step 5: Calculate \( 2B' \)**

\[ 2B' = 2 \cdot \begin{pmatrix} -1 & 2 \\ 3 & -4 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ 6 & -8 \end{pmatrix} \]

**Step 6: Calculate \( A' + 2B' \)**

\[ A' + 2B' = \begin{pmatrix} 2 & -7 \\ 8 & 3 \end{pmatrix} + \begin{pmatrix} -2 & 4 \\ 6 & -8 \end{pmatrix} = \begin{pmatrix} 2 + (-2) & -7 + 4 \\ 8 + 6 & 3 + (-8) \end{pmatrix} = \begin{pmatrix} 0 & -3 \\ 14 & -5 \end{pmatrix} \]

**Step 7: Compare \( (A + 2B)' \) and \( A' + 2B' \)**

Since \( (A + 2B)' = A' + 2B' \), the second statement is correct.

### (iii) Show that \( (AB)' = B'A' \)

**Step 1: Calculate \( AB \)**

\[ AB = \begin{pmatrix} 2 & 8 \\ -7 & 3 \end{pmatrix} \begin{pmatrix} -1 & 3 \\ 2 & -4 \end{pmatrix} = \begin{pmatrix} 2 \cdot (-1) + 8 \cdot 2 & 2 \cdot 3 + 8 \cdot (-4) \\ -7 \cdot (-1) + 3 \cdot 2 & -7 \cdot 3 + 3 \cdot (-4) \end{pmatrix} \]

Calculating the elements:

\[ AB = \begin{pmatrix} -2 + 16 & 6 - 32 \\ 7 + 6 & -21 - 12 \end{pmatrix} = \begin{pmatrix} 14 & -26 \\ 13 & -33 \end{pmatrix} \]

**Step 2: Calculate \( (AB)' \)**

\[ (AB)' = \begin{pmatrix} 14 & -26 \\ 13 & -33 \end{pmatrix}' = \begin{pmatrix} 14 & 13 \\ -26 & -33 \end{pmatrix} \]

**Step 3: Calculate \( B' \) and \( A' \)**

We already calculated:
\[ B' = \begin{pmatrix} -1 & 2 \\ 3 & -4 \end{pmatrix}, \quad A' = \begin{pmatrix} 2 & -7 \\ 8 & 3 \end{pmatrix} \]

**Step 4: Calculate \( B'A' \)**

\[ B'A' = \begin{pmatrix} -1 & 2 \\ 3 & -4 \end{pmatrix} \begin{pmatrix} 2 & -7 \\ 8 & 3 \end{pmatrix} = \begin{pmatrix} -1 \cdot 2 + 2 \cdot 8 & -1 \cdot (-7) + 2 \cdot 3 \\ 3 \cdot 2 + (-4) \cdot 8 & 3 \cdot (-7) + (-4) \cdot 3 \end{pmatrix} \]

Calculating the elements:

\[ B'A' = \begin{pmatrix} -2 + 16 & 7 + 6 \\ 6 - 32 & -21 - 12 \end{pmatrix} = \begin{pmatrix} 14 & 13 \\ -26 & -33 \end{pmatrix} \]

**Step 5: Compare \( (AB)' \) and \( B'A' \)**

Since \( (AB)' = B'A' \), the third statement is correct.

### Summary of Results:
1. \( (A + B)' \neq A + B' \) (Incorrect)
2. \( (A + 2B)' = A' + 2B' \) (Correct)
3. \( (AB)' = B'A' \) (Correct)