`if A=[{:(2,-3),(4,-1):}]and B=[{:(3,0),(-1,2):}],` then find matrix C such that 2A-B+3c is a unit matrix.

Explanation

To solve the problem, we need to find the matrix \( C \) such that the equation \( 2A - B + 3C = I \) holds true, where \( I \) is the identity matrix.

### Step-by-Step Solution:

1. **Identify the Matrices**:
Given:
\[ A = \begin{pmatrix} 2 & -3 \\ 4 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & 0 \\ -1 & 2 \end{pmatrix} \]

2. **Calculate \( 2A \)**:
Multiply each element of matrix \( A \) by 2:
\[ 2A = 2 \times \begin{pmatrix} 2 & -3 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} 4 & -6 \\ 8 & -2 \end{pmatrix} \]

3. **Calculate \( -B \)**:
Take the negative of matrix \( B \):
\[ -B = -\begin{pmatrix} 3 & 0 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 1 & -2 \end{pmatrix} \]

4. **Combine \( 2A \) and \( -B \)**:
Now, we add \( 2A \) and \( -B \):
\[ 2A - B = \begin{pmatrix} 4 & -6 \\ 8 & -2 \end{pmatrix} + \begin{pmatrix} -3 & 0 \\ 1 & -2 \end{pmatrix} = \begin{pmatrix} 4 - 3 & -6 + 0 \\ 8 + 1 & -2 - 2 \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 9 & -4 \end{pmatrix} \]

5. **Set Up the Equation**:
We have:
\[ 2A - B + 3C = I \]
Where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Thus, we can rewrite the equation as:
\[ \begin{pmatrix} 1 & -6 \\ 9 & -4 \end{pmatrix} + 3C = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

6. **Isolate \( 3C \)**:
Subtract \( \begin{pmatrix} 1 & -6 \\ 9 & -4 \end{pmatrix} \) from both sides:
\[ 3C = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & -6 \\ 9 & -4 \end{pmatrix} = \begin{pmatrix} 1 - 1 & 0 - (-6) \\ 0 - 9 & 1 - (-4) \end{pmatrix} = \begin{pmatrix} 0 & 6 \\ -9 & 5 \end{pmatrix} \]

7. **Solve for \( C \)**:
Divide each element of \( 3C \) by 3:
\[ C = \frac{1}{3} \begin{pmatrix} 0 & 6 \\ -9 & 5 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ -3 & \frac{5}{3} \end{pmatrix} \]

### Final Answer:
\[ C = \begin{pmatrix} 0 & 2 \\ -3 & \frac{5}{3} \end{pmatrix} \]