If `A=[{:(,1,2),(,3,4):}], B=[{:(,6,1),(,1,1):}] and C=[{:(,-2,-3),(,0,1):}]`. Find each of the following and state if they are equal:
(i) CA+B (ii) A+CB

Answer

To solve the problem, we need to find the matrices \( CA + B \) and \( A + CB \) and then check if they are equal.

Given matrices:
\[ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 6 & 1 \\ 1 & 1 \end{pmatrix}, \quad C = \begin{pmatrix} -2 & -3 \\ 0 & 1 \end{pmatrix} \]

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

To find \( CA \), we multiply matrix \( C \) with matrix \( A \):
\[ CA = \begin{pmatrix} -2 & -3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \]

Calculating the elements:
- First row, first column:
\[ (-2 \times 1) + (-3 \times 3) = -2 - 9 = -11 \]
- First row, second column:
\[ (-2 \times 2) + (-3 \times 4) = -4 - 12 = -16 \]
- Second row, first column:
\[ (0 \times 1) + (1 \times 3) = 0 + 3 = 3 \]
- Second row, second column:
\[ (0 \times 2) + (1 \times 4) = 0 + 4 = 4 \]

Thus,
\[ CA = \begin{pmatrix} -11 & -16 \\ 3 & 4 \end{pmatrix} \]

### Step 2: Calculate \( CA + B \)

Now, we add \( CA \) and \( B \):
\[ CA + B = \begin{pmatrix} -11 & -16 \\ 3 & 4 \end{pmatrix} + \begin{pmatrix} 6 & 1 \\ 1 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column:
\[ -11 + 6 = -5 \]
- First row, second column:
\[ -16 + 1 = -15 \]
- Second row, first column:
\[ 3 + 1 = 4 \]
- Second row, second column:
\[ 4 + 1 = 5 \]

Thus,
\[ CA + B = \begin{pmatrix} -5 & -15 \\ 4 & 5 \end{pmatrix} \]

### Step 3: Calculate \( CB \)

Next, we calculate \( CB \):
\[ CB = \begin{pmatrix} -2 & -3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 6 & 1 \\ 1 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column:
\[ (-2 \times 6) + (-3 \times 1) = -12 - 3 = -15 \]
- First row, second column:
\[ (-2 \times 1) + (-3 \times 1) = -2 - 3 = -5 \]
- Second row, first column:
\[ (0 \times 6) + (1 \times 1) = 0 + 1 = 1 \]
- Second row, second column:
\[ (0 \times 1) + (1 \times 1) = 0 + 1 = 1 \]

Thus,
\[ CB = \begin{pmatrix} -15 & -5 \\ 1 & 1 \end{pmatrix} \]

### Step 4: Calculate \( A + CB \)

Now, we add \( A \) and \( CB \):
\[ A + CB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} + \begin{pmatrix} -15 & -5 \\ 1 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column:
\[ 1 - 15 = -14 \]
- First row, second column:
\[ 2 - 5 = -3 \]
- Second row, first column:
\[ 3 + 1 = 4 \]
- Second row, second column:
\[ 4 + 1 = 5 \]

Thus,
\[ A + CB = \begin{pmatrix} -14 & -3 \\ 4 & 5 \end{pmatrix} \]

### Step 5: Compare \( CA + B \) and \( A + CB \)

Now we compare the two results:
\[ CA + B = \begin{pmatrix} -5 & -15 \\ 4 & 5 \end{pmatrix}, \quad A + CB = \begin{pmatrix} -14 & -3 \\ 4 & 5 \end{pmatrix} \]

Since the matrices are not equal, we conclude:
\[ CA + B \neq A + CB \]

### Summary of Results
- \( CA + B = \begin{pmatrix} -5 & -15 \\ 4 & 5 \end{pmatrix} \)
- \( A + CB = \begin{pmatrix} -14 & -3 \\ 4 & 5 \end{pmatrix} \)
- **Conclusion**: \( CA + B \neq A + CB \)