Let A+B+C=`[{:(,4,-1),(,0,1):}],4A+2B+C=[{:(,0,-1),(,-3,2):}]and 9A+3B+C=[{:(,0,2),(,2,1):}]"then find A"`

Answer

To solve the problem, we have three equations involving matrices A, B, and C. We need to find the matrix A given the equations:

1. \( A + B + C = \begin{pmatrix} 4 & -1 \\ 0 & 1 \end{pmatrix} \)
2. \( 4A + 2B + C = \begin{pmatrix} 0 & -1 \\ -3 & 2 \end{pmatrix} \)
3. \( 9A + 3B + C = \begin{pmatrix} 0 & 2 \\ 2 & 1 \end{pmatrix} \)

Let's denote:
- \( X = \begin{pmatrix} 4 & -1 \\ 0 & 1 \end{pmatrix} \)
- \( Y = \begin{pmatrix} 0 & -1 \\ -3 & 2 \end{pmatrix} \)
- \( Z = \begin{pmatrix} 0 & 2 \\ 2 & 1 \end{pmatrix} \)

### Step 1: Set up the equations

We have the following equations:
1. \( A + B + C = X \) (Equation 1)
2. \( 4A + 2B + C = Y \) (Equation 2)
3. \( 9A + 3B + C = Z \) (Equation 3)

### Step 2: Subtract Equation 1 from Equation 2

Subtracting Equation 1 from Equation 2:
\[ (4A + 2B + C) - (A + B + C) = Y - X \]
This simplifies to:
\[ 3A + B = Y - X \]

Calculating \( Y - X \):
\[ Y - X = \begin{pmatrix} 0 & -1 \\ -3 & 2 \end{pmatrix} - \begin{pmatrix} 4 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 - 4 & -1 + 1 \\ -3 - 0 & 2 - 1 \end{pmatrix} = \begin{pmatrix} -4 & 0 \\ -3 & 1 \end{pmatrix} \]
Thus, we have:
\[ 3A + B = \begin{pmatrix} -4 & 0 \\ -3 & 1 \end{pmatrix} \quad (Equation 4) \]

### Step 3: Subtract Equation 2 from Equation 3

Now, subtracting Equation 2 from Equation 3:
\[ (9A + 3B + C) - (4A + 2B + C) = Z - Y \]
This simplifies to:
\[ 5A + B = Z - Y \]

Calculating \( Z - Y \):
\[ Z - Y = \begin{pmatrix} 0 & 2 \\ 2 & 1 \end{pmatrix} - \begin{pmatrix} 0 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 0 - 0 & 2 + 1 \\ 2 + 3 & 1 - 2 \end{pmatrix} = \begin{pmatrix} 0 & 3 \\ 5 & -1 \end{pmatrix} \]
Thus, we have:
\[ 5A + B = \begin{pmatrix} 0 & 3 \\ 5 & -1 \end{pmatrix} \quad (Equation 5) \]

### Step 4: Subtract Equation 4 from Equation 5

Now, subtract Equation 4 from Equation 5:
\[ (5A + B) - (3A + B) = \begin{pmatrix} 0 & 3 \\ 5 & -1 \end{pmatrix} - \begin{pmatrix} -4 & 0 \\ -3 & 1 \end{pmatrix} \]
This simplifies to:
\[ 2A = \begin{pmatrix} 0 + 4 & 3 - 0 \\ 5 + 3 & -1 - 1 \end{pmatrix} = \begin{pmatrix} 4 & 3 \\ 8 & -2 \end{pmatrix} \]

### Step 5: Solve for A

Now, divide both sides by 2:
\[ A = \frac{1}{2} \begin{pmatrix} 4 & 3 \\ 8 & -2 \end{pmatrix} = \begin{pmatrix} 2 & \frac{3}{2} \\ 4 & -1 \end{pmatrix} \]

### Final Answer
Thus, the matrix \( A \) is:
\[ A = \begin{pmatrix} 2 & \frac{3}{2} \\ 4 & -1 \end{pmatrix} \]