Solve the following system of equation by matrix method `x + y - 2z = 1; x - y + z = 0; 2x + 3y - 4z = 2,`

Answer

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

### Given Equations:
1. \( x + y - 2z = 1 \) (Equation 1)
2. \( x - y + z = 0 \) (Equation 2)
3. \( 2x + 3y - 4z = 2 \) (Equation 3)

### Step 1: Form the Coefficient Matrix and the Constant Matrix
We can represent the system of equations in matrix form as \( AX = B \), where:

- \( A \) is the coefficient matrix:
\[ A = \begin{bmatrix} 1 & 1 & -2 \\ 1 & -1 & 1 \\ 2 & 3 & -4 \end{bmatrix} \]

- \( X \) is the column matrix of variables:
\[ X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

- \( B \) is the constant matrix:
\[ B = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \]

### Step 2: Calculate the Determinant of Matrix A (denoted as \( \Delta \))
To find \( \Delta \), we compute the determinant of matrix \( A \):
\[ \Delta = \begin{vmatrix} 1 & 1 & -2 \\ 1 & -1 & 1 \\ 2 & 3 & -4 \end{vmatrix} \]

Calculating the determinant:
\[ \Delta = 1 \cdot \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 2 & -4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} = (-1)(-4) - (1)(3) = 4 - 3 = 1 \)

2. \( \begin{vmatrix} 1 & 1 \\ 2 & -4 \end{vmatrix} = (1)(-4) - (1)(2) = -4 - 2 = -6 \)

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

Substituting these values back:
\[ \Delta = 1 \cdot 1 - 1 \cdot (-6) - 2 \cdot 5 = 1 + 6 - 10 = -3 \]

### Step 3: Calculate \( \Delta_1 \), \( \Delta_2 \), and \( \Delta_3 \)
#### Finding \( \Delta_1 \)
Replace the first column of \( A \) with the constants from \( B \):
\[ \Delta_1 = \begin{vmatrix} 1 & 1 & -2 \\ 0 & -1 & 1 \\ 2 & 3 & -4 \end{vmatrix} \]

Calculating \( \Delta_1 \):
\[ \Delta_1 = 1 \cdot \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 1 \\ 2 & -4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & -1 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 1 \\ 3 & -4 \end{vmatrix} = 1 \) (as calculated earlier)

2. \( \begin{vmatrix} 0 & 1 \\ 2 & -4 \end{vmatrix} = (0)(-4) - (1)(2) = -2 \)

3. \( \begin{vmatrix} 0 & -1 \\ 2 & 3 \end{vmatrix} = (0)(3) - (-1)(2) = 2 \)

Substituting these values back:
\[ \Delta_1 = 1 \cdot 1 - 1 \cdot (-2) - 2 \cdot 2 = 1 + 2 - 4 = -1 \]

#### Finding \( \Delta_2 \)
Replace the second column of \( A \) with the constants from \( B \):
\[ \Delta_2 = \begin{vmatrix} 1 & 1 & -2 \\ 1 & 0 & 1 \\ 2 & 2 & -4 \end{vmatrix} \]

Calculating \( \Delta_2 \):
\[ \Delta_2 = 1 \cdot \begin{vmatrix} 0 & 1 \\ 2 & -4 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 2 & -4 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 0 \\ 2 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 0 & 1 \\ 2 & -4 \end{vmatrix} = -2 \)

2. \( \begin{vmatrix} 1 & 1 \\ 2 & -4 \end{vmatrix} = -6 \)

3. \( \begin{vmatrix} 1 & 0 \\ 2 & 2 \end{vmatrix} = 2 \)

Substituting these values back:
\[ \Delta_2 = 1 \cdot (-2) - 1 \cdot (-6) - 2 \cdot 2 = -2 + 6 - 4 = 0 \]

#### Finding \( \Delta_3 \)
Replace the third column of \( A \) with the constants from \( B \):
\[ \Delta_3 = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 2 & 3 & 2 \end{vmatrix} \]

Calculating \( \Delta_3 \):
\[ \Delta_3 = 1 \cdot \begin{vmatrix} -1 & 0 \\ 3 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 0 \\ 3 & 2 \end{vmatrix} = -2 \)

2. \( \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} = 0 \)

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

Substituting these values back:
\[ \Delta_3 = 1 \cdot (-2) - 1 \cdot 0 + 1 \cdot 5 = -2 + 0 + 5 = 3 \]

### Step 4: Calculate the Values of \( x \), \( y \), and \( z \)
Using Cramer's rule:
\[ x = \frac{\Delta_1}{\Delta} = \frac{-1}{-3} = \frac{1}{3} \]
\[ y = \frac{\Delta_2}{\Delta} = \frac{0}{-3} = 0 \]
\[ z = \frac{\Delta_3}{\Delta} = \frac{3}{-3} = -1 \]

### Final Solution
The solution to the system of equations is:
\[ x = \frac{1}{3}, \quad y = 0, \quad z = -1 \]