To solve the given matrices problem, we will follow these steps:
### Given Matrices
- \( A = \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} \)
- \( B = \begin{pmatrix} 1 & 1 \\ 5 & 2 \end{pmatrix} \)
- \( C = \begin{pmatrix} -3 & -1 \\ 0 & 0 \end{pmatrix} \)
### Part (i): Find \( 2A - 3B + C \)
1. **Calculate \( 2A \)**:
\[
2A = 2 \times \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 6 & 0 \end{pmatrix}
\]
**Hint**: Multiply each element of matrix \( A \) by 2.
2. **Calculate \( -3B \)**:
\[
-3B = -3 \times \begin{pmatrix} 1 & 1 \\ 5 & 2 \end{pmatrix} = \begin{pmatrix} -3 & -3 \\ -15 & -6 \end{pmatrix}
\]
**Hint**: Multiply each element of matrix \( B \) by -3.
3. **Add \( C \)**:
\[
C = \begin{pmatrix} -3 & -1 \\ 0 & 0 \end{pmatrix}
\]
4. **Combine all matrices**:
\[
2A - 3B + C = \begin{pmatrix} 4 & 2 \\ 6 & 0 \end{pmatrix} + \begin{pmatrix} -3 & -3 \\ -15 & -6 \end{pmatrix} + \begin{pmatrix} -3 & -1 \\ 0 & 0 \end{pmatrix}
\]
Now, add the matrices element-wise:
- First row, first column: \( 4 - 3 - 3 = -2 \)
- First row, second column: \( 2 - 3 - 1 = -2 \)
- Second row, first column: \( 6 - 15 + 0 = -9 \)
- Second row, second column: \( 0 - 6 + 0 = -6 \)
Thus,
\[
2A - 3B + C = \begin{pmatrix} -2 & -2 \\ -9 & -6 \end{pmatrix}
\]
### Part (ii): Find \( A + 2C - B \)
1. **Calculate \( 2C \)**:
\[
2C = 2 \times \begin{pmatrix} -3 & -1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} -6 & -2 \\ 0 & 0 \end{pmatrix}
\]
**Hint**: Multiply each element of matrix \( C \) by 2.
2. **Combine \( A \), \( 2C \), and \( -B \)**:
\[
A + 2C - B = \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix} + \begin{pmatrix} -6 & -2 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 1 \\ 5 & 2 \end{pmatrix}
\]
First, we can add \( A \) and \( 2C \):
- First row, first column: \( 2 - 6 = -4 \)
- First row, second column: \( 1 - 2 = -1 \)
- Second row, first column: \( 3 + 0 = 3 \)
- Second row, second column: \( 0 + 0 = 0 \)
So,
\[
A + 2C = \begin{pmatrix} -4 & -1 \\ 3 & 0 \end{pmatrix}
\]
3. **Now subtract \( B \)**:
\[
A + 2C - B = \begin{pmatrix} -4 & -1 \\ 3 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 1 \\ 5 & 2 \end{pmatrix}
\]
Now, subtract element-wise:
- First row, first column: \( -4 - 1 = -5 \)
- First row, second column: \( -1 - 1 = -2 \)
- Second row, first column: \( 3 - 5 = -2 \)
- Second row, second column: \( 0 - 2 = -2 \)
Thus,
\[
A + 2C - B = \begin{pmatrix} -5 & -2 \\ -2 & -2 \end{pmatrix}
\]
### Final Answers:
(i) \( 2A - 3B + C = \begin{pmatrix} -2 & -2 \\ -9 & -6 \end{pmatrix} \)
(ii) \( A + 2C - B = \begin{pmatrix} -5 & -2 \\ -2 & -2 \end{pmatrix} \)
