2x+3y+3z=5
x-2y+z=-4
3x-y-2z=3

Answer

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

### Given Equations:
1. \( 2x + 3y + 3z = 5 \) (Equation 1)
2. \( x - 2y + z = -4 \) (Equation 2)
3. \( 3x - y - 2z = 3 \) (Equation 3)

### Step 1: Write the system in matrix form
We can express 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.

The coefficient matrix \( A \) and the constant matrix \( B \) are given by:
\[ A = \begin{pmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix} \]

### Step 2: Find the inverse of matrix \( A \)
To find \( X \), we need to calculate \( A^{-1} \). We will first find the determinant of \( A \).

#### Determinant of \( A \):
\[ \text{det}(A) = 2 \begin{vmatrix} -2 & 1 \\ -1 & -2 \end{vmatrix} - 3 \begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} + 3 \begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} \]
Calculating the minors:
1. \( \begin{vmatrix} -2 & 1 \\ -1 & -2 \end{vmatrix} = (-2)(-2) - (1)(-1) = 4 + 1 = 5 \)
2. \( \begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} = (1)(-2) - (1)(3) = -2 - 3 = -5 \)
3. \( \begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (-2)(3) = -1 + 6 = 5 \)

Putting it all together:
\[ \text{det}(A) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 \]

### Step 3: Find the adjoint of \( A \)
The adjoint of \( A \) is the transpose of the cofactor matrix. We will calculate the cofactors and then transpose.

Calculating cofactors:
1. \( C_{11} = \begin{vmatrix} -2 & 1 \\ -1 & -2 \end{vmatrix} = 5 \) (cofactor is \( 5 \))
2. \( C_{12} = -\begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} = 5 \) (cofactor is \( -5 \))
3. \( C_{13} = \begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} = 5 \) (cofactor is \( 5 \))
4. \( C_{21} = -\begin{vmatrix} 3 & 3 \\ -1 & -2 \end{vmatrix} = -3 \) (cofactor is \( 3 \))
5. \( C_{22} = \begin{vmatrix} 2 & 3 \\ 3 & -2 \end{vmatrix} = -12 \) (cofactor is \( -12 \))
6. \( C_{23} = -\begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = 9 \) (cofactor is \( -9 \))
7. \( C_{31} = \begin{vmatrix} 3 & 3 \\ -2 & 1 \end{vmatrix} = 9 \) (cofactor is \( 9 \))
8. \( C_{32} = -\begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = -3 \) (cofactor is \( 3 \))
9. \( C_{33} = \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} = -7 \) (cofactor is \( -7 \))

The cofactor matrix is:
\[ \text{Cofactor}(A) = \begin{pmatrix} 5 & -5 & 5 \\ 3 & -12 & 9 \\ 9 & 3 & -7 \end{pmatrix} \]

Taking the transpose gives us the adjoint:
\[ \text{adj}(A) = \begin{pmatrix} 5 & 3 & 9 \\ -5 & -12 & 3 \\ 5 & 9 & -7 \end{pmatrix} \]

### Step 4: Calculate the inverse of \( A \)
Using the formula \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \):
\[ A^{-1} = \frac{1}{40} \begin{pmatrix} 5 & 3 & 9 \\ -5 & -12 & 3 \\ 5 & 9 & -7 \end{pmatrix} \]

### Step 5: Solve for \( X \)
Now, we can find \( X \) using \( X = A^{-1}B \):
\[ X = \frac{1}{40} \begin{pmatrix} 5 & 3 & 9 \\ -5 & -12 & 3 \\ 5 & 9 & -7 \end{pmatrix} \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix} \]

Calculating the product:
1. First row: \( 5(5) + 3(-4) + 9(3) = 25 - 12 + 27 = 40 \)
2. Second row: \( -5(5) + -12(-4) + 3(3) = -25 + 48 + 9 = 32 \)
3. Third row: \( 5(5) + 9(-4) + -7(3) = 25 - 36 - 21 = -32 \)

Thus,
\[ X = \frac{1}{40} \begin{pmatrix} 40 \\ 32 \\ -32 \end{pmatrix} = \begin{pmatrix} 1 \\ 0.8 \\ -0.8 \end{pmatrix} \]

### Final Solution:
The solution to the system of equations is:
\[ x = 1, \quad y = 0.8, \quad z = -0.8 \]