Using matrix method, solve the system of equation `3x+2y-2z=3, x+2y+3z=6` and `2x-y+z=2`

Explanation

To solve the system of equations using the matrix method, we will follow these steps:

### Given Equations:
1. \( 3x + 2y - 2z = 3 \) (Equation 1)
2. \( x + 2y + 3z = 6 \) (Equation 2)
3. \( 2x - y + z = 2 \) (Equation 3)

### Step 1: Formulate the Matrices
We can represent the system of equations in the form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the variable matrix,
- \( B \) is the constant matrix.

#### Coefficient Matrix \( A \):
\[ A = \begin{bmatrix} 3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{bmatrix} \]

#### Variable Matrix \( X \):
\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

#### Constant Matrix \( B \):
\[ B = \begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix} \]

### Step 2: Find the Determinant of Matrix \( A \)
To find the inverse of matrix \( A \), we first need to calculate its determinant \( |A| \).

\[ |A| = 3 \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} - 2 \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} - 2 \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} = (2)(1) - (3)(-1) = 2 + 3 = 5 \)

2. \( \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = (1)(1) - (3)(2) = 1 - 6 = -5 \)

3. \( \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \)

Now substituting back into the determinant formula:
\[ |A| = 3(5) - 2(-5) - 2(-5) = 15 + 10 + 10 = 35 \]

### Step 3: Find the Inverse of Matrix \( A \)
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{|A|} \text{Adj}(A) \]

To find the adjoint of \( A \), we need to calculate the cofactor matrix and then transpose it.

#### Cofactor Matrix Calculation:
1. \( C_{11} = \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} = 5 \)
2. \( C_{12} = -\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 5 \)
3. \( C_{13} = \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = -5 \)

4. \( C_{21} = -\begin{vmatrix} 2 & -2 \\ -1 & 1 \end{vmatrix} = 2 \)
5. \( C_{22} = \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = 7 \)
6. \( C_{23} = -\begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = 7 \)

7. \( C_{31} = \begin{vmatrix} 2 & -2 \\ 2 & 3 \end{vmatrix} = 6 \)
8. \( C_{32} = -\begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = 11 \)
9. \( C_{33} = \begin{vmatrix} 3 & 2 \\ 1 & 2 \end{vmatrix} = 4 \)

The cofactor matrix is:
\[ C = \begin{bmatrix} 5 & 5 & -5 \\ 2 & 7 & 7 \\ 6 & 11 & 4 \end{bmatrix} \]

Transposing the cofactor matrix gives us the adjoint:
\[ \text{Adj}(A) = \begin{bmatrix} 5 & 2 & 6 \\ 5 & 7 & 11 \\ -5 & 7 & 4 \end{bmatrix} \]

Now, we can find the inverse:
\[ A^{-1} = \frac{1}{35} \begin{bmatrix} 5 & 2 & 6 \\ 5 & 7 & 11 \\ -5 & 7 & 4 \end{bmatrix} \]

### Step 4: Solve for \( X \)
Now we can find \( X \) using:
\[ X = A^{-1}B \]

Calculating \( A^{-1}B \):
\[ X = \frac{1}{35} \begin{bmatrix} 5 & 2 & 6 \\ 5 & 7 & 11 \\ -5 & 7 & 4 \end{bmatrix} \begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix} \]

Calculating the product:
1. First row: \( 5(3) + 2(6) + 6(2) = 15 + 12 + 12 = 39 \)
2. Second row: \( 5(3) + 7(6) + 11(2) = 15 + 42 + 22 = 79 \)
3. Third row: \( -5(3) + 7(6) + 4(2) = -15 + 42 + 8 = 35 \)

Thus,
\[ X = \frac{1}{35} \begin{bmatrix} 39 \\ 79 \\ 35 \end{bmatrix} = \begin{bmatrix} \frac{39}{35} \\ \frac{79}{35} \\ 1 \end{bmatrix} \]

### Final Solution
The solution to the system of equations is:
\[ x = \frac{39}{35}, \quad y = \frac{79}{35}, \quad z = 1 \]