`if A=[{:(2,1,3),(1,-1,2),(4,1,5):}]and B=[{:(1,-1,2),(2,1,5),(4,1,3):}],`then show that : `(i) (A+B)'=A'+B'``(ii) (A+4B)'=A'+4B'`

Explanation

To solve the given problem, we need to show two properties of matrix transposition involving matrices A and B. Let's go through each part step by step.

### Given Matrices:
Let
\[ A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & -1 & 2 \\ 4 & 1 & 5 \end{pmatrix} \]
and
\[ B = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 5 \\ 4 & 1 & 3 \end{pmatrix} \]

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

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

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

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

Calculating each element:

\[ A + B = \begin{pmatrix} 3 & 0 & 5 \\ 3 & 0 & 7 \\ 8 & 2 & 8 \end{pmatrix} \]

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

Now we find the transpose of \( A + B \):

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

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

Now we calculate the transposes of A and B individually.

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

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

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

Now we add \( A' \) and \( B' \):

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

Calculating each element:

\[ A' + B' = \begin{pmatrix} 3 & 3 & 8 \\ 0 & 0 & 2 \\ 5 & 7 & 8 \end{pmatrix} \]

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

We see that:

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

Thus, we have shown that \( (A + B)' = A' + B' \).

### Part (ii): Show that \( (A + 4B)' = A' + 4B' \)

**Step 1: Calculate \( 4B \)**

First, we multiply matrix B by 4:

\[ 4B = 4 \times \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 5 \\ 4 & 1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & -4 & 8 \\ 8 & 4 & 20 \\ 16 & 4 & 12 \end{pmatrix} \]

**Step 2: Calculate \( A + 4B \)**

Now we add A and \( 4B \):

\[ A + 4B = \begin{pmatrix} 2 + 4 & 1 - 4 & 3 + 8 \\ 1 + 8 & -1 + 4 & 2 + 20 \\ 4 + 16 & 1 + 4 & 5 + 12 \end{pmatrix} \]

Calculating each element:

\[ A + 4B = \begin{pmatrix} 6 & -3 & 11 \\ 9 & 3 & 22 \\ 20 & 5 & 17 \end{pmatrix} \]

**Step 3: Calculate \( (A + 4B)' \)**

Now we find the transpose of \( A + 4B \):

\[ (A + 4B)' = \begin{pmatrix} 6 & 9 & 20 \\ -3 & 3 & 5 \\ 11 & 22 & 17 \end{pmatrix} \]

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

Now we calculate the transpose of B and then multiply by 4:

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

\[ 4B' = 4 \times \begin{pmatrix} 1 & 2 & 4 \\ -1 & 1 & 1 \\ 2 & 5 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 8 & 16 \\ -4 & 4 & 4 \\ 8 & 20 & 12 \end{pmatrix} \]

**Step 5: Calculate \( A' + 4B' \)**

Now we add \( A' \) and \( 4B' \):

\[ A' + 4B' = \begin{pmatrix} 2 + 4 & 1 + 8 & 4 + 16 \\ 1 - 4 & -1 + 4 & 1 + 4 \\ 3 + 8 & 2 + 20 & 5 + 12 \end{pmatrix} \]

Calculating each element:

\[ A' + 4B' = \begin{pmatrix} 6 & 9 & 20 \\ -3 & 3 & 5 \\ 11 & 22 & 17 \end{pmatrix} \]

**Step 6: Compare \( (A + 4B)' \) and \( A' + 4B' \)**

We see that:

\[ (A + 4B)' = \begin{pmatrix} 6 & 9 & 20 \\ -3 & 3 & 5 \\ 11 & 22 & 17 \end{pmatrix} = A' + 4B' \]

Thus, we have shown that \( (A + 4B)' = A' + 4B' \).

### Final Conclusion:
Both parts of the problem have been proven:
1. \( (A + B)' = A' + B' \)
2. \( (A + 4B)' = A' + 4B' \)