2x+8y+5z=5
x+y+z=-2
x+2y-z=2

Answer

To solve the system of equations using the Gauss-Jordan method, we will follow these steps:

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( 2x + 8y + 5z = 5 \)
2. \( x + y + z = -2 \)
3. \( x + 2y - z = 2 \)

We can represent this system in matrix form as \( AX = B \), where:
\[ A = \begin{bmatrix} 2 & 8 & 5 \\ 1 & 1 & 1 \\ 1 & 2 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ -2 \\ 2 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix A
The determinant of matrix \( A \) can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \text{det}(A) = 2 \cdot (1 \cdot (-1) - 1 \cdot 2) - 8 \cdot (1 \cdot (-1) - 1 \cdot 1) + 5 \cdot (1 \cdot 2 - 1 \cdot 1) \]
Calculating this gives:
\[ = 2 \cdot (-1 - 2) - 8 \cdot (-1 - 1) + 5 \cdot (2 - 1) = 2 \cdot (-3) - 8 \cdot (-2) + 5 \cdot 1 = -6 + 16 + 5 = 15 \]

### Step 3: Find the adjoint of matrix A
To find the adjoint, we first need to calculate the cofactor matrix and then take its transpose.

1. **Cofactor matrix**:
- \( C_{11} = \text{det} \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix} = (1)(-1) - (1)(2) = -1 - 2 = -3 \)
- \( C_{12} = -\text{det} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = -((1)(-1) - (1)(1)) = -(-1 - 1) = 2 \)
- \( C_{13} = \text{det} \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1 \)
- \( C_{21} = -\text{det} \begin{bmatrix} 8 & 5 \\ 2 & -1 \end{bmatrix} = -((8)(-1) - (5)(2)) = -(-8 - 10) = 18 \)
- \( C_{22} = \text{det} \begin{bmatrix} 2 & 5 \\ 1 & -1 \end{bmatrix} = (2)(-1) - (5)(1) = -2 - 5 = -7 \)
- \( C_{23} = -\text{det} \begin{bmatrix} 2 & 8 \\ 1 & 1 \end{bmatrix} = -((2)(1) - (8)(1)) = -2 + 8 = 6 \)
- \( C_{31} = \text{det} \begin{bmatrix} 8 & 5 \\ 1 & 1 \end{bmatrix} = (8)(1) - (5)(1) = 8 - 5 = 3 \)
- \( C_{32} = -\text{det} \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix} = -((2)(1) - (5)(1)) = -2 + 5 = 3 \)
- \( C_{33} = \text{det} \begin{bmatrix} 2 & 8 \\ 1 & 1 \end{bmatrix} = (2)(1) - (8)(1) = 2 - 8 = -6 \)

Thus, the cofactor matrix is:
\[ C = \begin{bmatrix} -3 & 2 & 1 \\ 18 & -7 & 6 \\ 3 & 3 & -6 \end{bmatrix} \]

2. **Adjoint matrix** (transpose of cofactor matrix):
\[ \text{adj}(A) = \begin{bmatrix} -3 & 18 & 3 \\ 2 & -7 & 3 \\ 1 & 6 & -6 \end{bmatrix} \]

### Step 4: Calculate the solution using the formula \( X = A^{-1}B \)
Using the formula \( A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \):
\[ X = \frac{1}{15} \begin{bmatrix} -3 & 18 & 3 \\ 2 & -7 & 3 \\ 1 & 6 & -6 \end{bmatrix} \begin{bmatrix} 5 \\ -2 \\ 2 \end{bmatrix} \]

Calculating the product:
\[ X = \frac{1}{15} \begin{bmatrix} -3 \cdot 5 + 18 \cdot (-2) + 3 \cdot 2 \\ 2 \cdot 5 + (-7) \cdot (-2) + 3 \cdot 2 \\ 1 \cdot 5 + 6 \cdot (-2) + (-6) \cdot 2 \end{bmatrix} \]
Calculating each component:
1. First row: \( -15 - 36 + 6 = -45 \)
2. Second row: \( 10 + 14 + 6 = 30 \)
3. Third row: \( 5 - 12 - 12 = -19 \)

Thus,
\[ X = \frac{1}{15} \begin{bmatrix} -45 \\ 30 \\ -19 \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \\ -1.267 \end{bmatrix} \]

### Final Solution
The solution to the system of equations is:
\[ x = -3, \quad y = 2, \quad z \approx -1.267 \]