Let `A=[(1,2),(2,1)],B=[(2,0),(1,3)]` and `C=[(1,1),(2,3)]`.
Calculate AC, BC and (A+B) C.
Also, verify that (A+B)C=AC+BC.

Answer

To solve the problem, we will calculate the matrices \( AC \), \( BC \), and \( (A + B)C \) step by step, and then verify that \( (A + B)C = AC + BC \).

Given matrices:
\[ A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}, \quad C = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \]

### Step 1: Calculate \( AC \)

To find \( AC \), we multiply matrix \( A \) by matrix \( C \):
\[ AC = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \]

Calculating each element:
- First row, first column: \( 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 \)
- First row, second column: \( 1 \cdot 1 + 2 \cdot 3 = 1 + 6 = 7 \)
- Second row, first column: \( 2 \cdot 1 + 1 \cdot 2 = 2 + 2 = 4 \)
- Second row, second column: \( 2 \cdot 1 + 1 \cdot 3 = 2 + 3 = 5 \)

Thus,
\[ AC = \begin{pmatrix} 5 & 7 \\ 4 & 5 \end{pmatrix} \]

### Step 2: Calculate \( BC \)

Next, we calculate \( BC \):
\[ BC = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \]

Calculating each element:
- First row, first column: \( 2 \cdot 1 + 0 \cdot 2 = 2 + 0 = 2 \)
- First row, second column: \( 2 \cdot 1 + 0 \cdot 3 = 2 + 0 = 2 \)
- Second row, first column: \( 1 \cdot 1 + 3 \cdot 2 = 1 + 6 = 7 \)
- Second row, second column: \( 1 \cdot 1 + 3 \cdot 3 = 1 + 9 = 10 \)

Thus,
\[ BC = \begin{pmatrix} 2 & 2 \\ 7 & 10 \end{pmatrix} \]

### Step 3: Calculate \( A + B \)

Now, we calculate \( A + B \):
\[ A + B = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} \]

Calculating each element:
- First row, first column: \( 1 + 2 = 3 \)
- First row, second column: \( 2 + 0 = 2 \)
- Second row, first column: \( 2 + 1 = 3 \)
- Second row, second column: \( 1 + 3 = 4 \)

Thus,
\[ A + B = \begin{pmatrix} 3 & 2 \\ 3 & 4 \end{pmatrix} \]

### Step 4: Calculate \( (A + B)C \)

Now we calculate \( (A + B)C \):
\[ (A + B)C = \begin{pmatrix} 3 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \]

Calculating each element:
- First row, first column: \( 3 \cdot 1 + 2 \cdot 2 = 3 + 4 = 7 \)
- First row, second column: \( 3 \cdot 1 + 2 \cdot 3 = 3 + 6 = 9 \)
- Second row, first column: \( 3 \cdot 1 + 4 \cdot 2 = 3 + 8 = 11 \)
- Second row, second column: \( 3 \cdot 1 + 4 \cdot 3 = 3 + 12 = 15 \)

Thus,
\[ (A + B)C = \begin{pmatrix} 7 & 9 \\ 11 & 15 \end{pmatrix} \]

### Step 5: Verify \( (A + B)C = AC + BC \)

Now we need to verify if \( (A + B)C = AC + BC \):
\[ AC + BC = \begin{pmatrix} 5 & 7 \\ 4 & 5 \end{pmatrix} + \begin{pmatrix} 2 & 2 \\ 7 & 10 \end{pmatrix} \]

Calculating each element:
- First row, first column: \( 5 + 2 = 7 \)
- First row, second column: \( 7 + 2 = 9 \)
- Second row, first column: \( 4 + 7 = 11 \)
- Second row, second column: \( 5 + 10 = 15 \)

Thus,
\[ AC + BC = \begin{pmatrix} 7 & 9 \\ 11 & 15 \end{pmatrix} \]

Since \( (A + B)C = AC + BC \), we have verified the equality.

### Final Result
\[ AC = \begin{pmatrix} 5 & 7 \\ 4 & 5 \end{pmatrix}, \quad BC = \begin{pmatrix} 2 & 2 \\ 7 & 10 \end{pmatrix}, \quad (A + B)C = \begin{pmatrix} 7 & 9 \\ 11 & 15 \end{pmatrix} \]