`if A=[{:(-1,2,,3),(5,7,9),(-2,1,1):}]and B=[{:(-4,1,-5),(1,2,0),(1,3,1):}],`then verify that `(I) (A+B)'=A'+B',(ii) (A-b)'=A'=B'`

Explanation

To verify the given statements about matrices \( A \) and \( B \), we will follow these steps:

### Given:
\[ A = \begin{pmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{pmatrix} \]
\[ B = \begin{pmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{pmatrix} \]

### Step 1: Calculate \( A + B \)
To find \( A + B \), we add the corresponding elements of matrices \( A \) and \( B \).

\[ A + B = \begin{pmatrix} -1 + (-4) & 2 + 1 & 3 + (-5) \\ 5 + 1 & 7 + 2 & 9 + 0 \\ -2 + 1 & 1 + 3 & 1 + 1 \end{pmatrix} \]

Calculating each element:
\[ A + B = \begin{pmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{pmatrix} \]

### Step 2: Calculate \( (A + B)' \) (Transpose of \( A + B \))
To find the transpose, we swap rows and columns.

\[ (A + B)' = \begin{pmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{pmatrix} \]

### Step 3: Calculate \( A' \) (Transpose of \( A \))
\[ A' = \begin{pmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{pmatrix} \]

### Step 4: Calculate \( B' \) (Transpose of \( B \))
\[ B' = \begin{pmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{pmatrix} \]

### Step 5: Calculate \( A' + B' \)
Now we add the transposed matrices \( A' \) and \( B' \).

\[ A' + B' = \begin{pmatrix} -1 + (-4) & 5 + 1 & -2 + 1 \\ 2 + 1 & 7 + 2 & 1 + 3 \\ 3 + (-5) & 9 + 0 & 1 + 1 \end{pmatrix} \]

Calculating each element:
\[ A' + B' = \begin{pmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{pmatrix} \]

### Step 6: Verify \( (A + B)' = A' + B' \)
Now we compare \( (A + B)' \) and \( A' + B' \):

\[ (A + B)' = \begin{pmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{pmatrix} \]
\[ A' + B' = \begin{pmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{pmatrix} \]

Since both matrices are equal, we have verified that:
\[ (A + B)' = A' + B' \]

### Step 7: Calculate \( A - B \)
Now we calculate \( A - B \).

\[ A - B = \begin{pmatrix} -1 - (-4) & 2 - 1 & 3 - (-5) \\ 5 - 1 & 7 - 2 & 9 - 0 \\ -2 - 1 & 1 - 3 & 1 - 1 \end{pmatrix} \]

Calculating each element:
\[ A - B = \begin{pmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{pmatrix} \]

### Step 8: Calculate \( (A - B)' \) (Transpose of \( A - B \))
\[ (A - B)' = \begin{pmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{pmatrix} \]

### Step 9: Calculate \( A' - B' \)
Now we subtract \( B' \) from \( A' \).

\[ A' - B' = \begin{pmatrix} -1 - (-4) & 5 - 1 & -2 - 1 \\ 2 - 1 & 7 - 2 & 1 - 3 \\ 3 - (-5) & 9 - 0 & 1 - 1 \end{pmatrix} \]

Calculating each element:
\[ A' - B' = \begin{pmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{pmatrix} \]

### Step 10: Verify \( (A - B)' = A' - B' \)
Now we compare \( (A - B)' \) and \( A' - B' \):

\[ (A - B)' = \begin{pmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{pmatrix} \]
\[ A' - B' = \begin{pmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{pmatrix} \]

Since both matrices are equal, we have verified that:
\[ (A - B)' = A' - B' \]

### Conclusion
Both parts of the verification are complete:
1. \( (A + B)' = A' + B' \)
2. \( (A - B)' = A' - B' \)

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