Solve the equations by matrix method: x+y+z=6 x+y-z=0 2x+y+2z=10

Answer

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

### Step 1: Write the equations in standard form
The equations given are:
1. \( x + y + z = 6 \)
2. \( x + y - z = 0 \)
3. \( 2x + y + 2z = 10 \)

### Step 2: Form the coefficient matrix \( A \), variable matrix \( X \), and constant matrix \( B \)
From the equations, we can identify:
- Coefficient matrix \( A \):
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 2 & 1 & 2 \end{bmatrix} \]

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

- Constant matrix \( B \):
\[ B = \begin{bmatrix} 6 \\ 0 \\ 10 \end{bmatrix} \]

### Step 3: Set up the matrix equation \( AX = B \)
The matrix equation can be written as:
\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 2 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 0 \\ 10 \end{bmatrix} \]

### Step 4: Find the determinant of matrix \( A \)
To find the inverse of matrix \( A \), we first need to calculate its determinant \( |A| \):
\[ |A| = 1 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \]

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

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

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

Now substituting back into the determinant formula:
\[ |A| = 1 \cdot 3 - 1 \cdot 4 + 1 \cdot (-1) = 3 - 4 - 1 = -2 \]

### Step 5: Find the inverse of matrix \( A \)
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) \]
Since \( |A| = -2 \), we need to find the adjugate of \( A \).

### Step 6: Calculate the adjugate of \( A \)
The adjugate matrix is formed by the cofactors of \( A \):
1. Calculate cofactors for each element of \( A \):
- \( C_{11} = \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} = 3 \)
- \( C_{12} = -\begin{vmatrix} 1 & -1 \\ 2 & 2 \end{vmatrix} = -4 \)
- \( C_{13} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -1 \)
- \( C_{21} = -\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = -1 \)
- \( C_{22} = \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = 0 \)
- \( C_{23} = -\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = 1 \)
- \( C_{31} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2 \)
- \( C_{32} = -\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = 2 \)
- \( C_{33} = \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 0 \)

The cofactor matrix is:
\[ \text{Cof}(A) = \begin{bmatrix} 3 & -4 & -1 \\ -1 & 0 & 1 \\ -2 & 2 & 0 \end{bmatrix} \]

Taking the transpose gives us the adjugate:
\[ \text{adj}(A) = \begin{bmatrix} 3 & -1 & -2 \\ -4 & 0 & 2 \\ -1 & 1 & 0 \end{bmatrix} \]

### Step 7: Calculate \( A^{-1} \)
Now we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{-2} \begin{bmatrix} 3 & -1 & -2 \\ -4 & 0 & 2 \\ -1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} -1.5 & 0.5 & 1 \\ 2 & 0 & -1 \\ 0.5 & -0.5 & 0 \end{bmatrix} \]

### Step 8: Multiply \( A^{-1} \) with \( B \) to find \( X \)
Now we can find \( X \):
\[ X = A^{-1}B = \begin{bmatrix} -1.5 & 0.5 & 1 \\ 2 & 0 & -1 \\ 0.5 & -0.5 & 0 \end{bmatrix} \begin{bmatrix} 6 \\ 0 \\ 10 \end{bmatrix} \]

Calculating the multiplication:
1. First row: \( -1.5 \cdot 6 + 0.5 \cdot 0 + 1 \cdot 10 = -9 + 0 + 10 = 1 \)
2. Second row: \( 2 \cdot 6 + 0 \cdot 0 - 1 \cdot 10 = 12 + 0 - 10 = 2 \)
3. Third row: \( 0.5 \cdot 6 - 0.5 \cdot 0 + 0 \cdot 10 = 3 - 0 + 0 = 3 \)

Thus, we have:
\[ X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]

### Final Solution
The solution to the system of equations is:
\[ x = 1, \quad y = 2, \quad z = 3 \]