x-2y+z=0
y-z=2
2x-3z=10

Answer

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

Given equations:
1. \( x - 2y + z = 0 \) (Equation 1)
2. \( y - z = 2 \) (Equation 2)
3. \( 2x - 3z = 10 \) (Equation 3)

### Step 1: Write the equations in standard form
We can rewrite the equations in standard form to identify the coefficients:
1. \( x - 2y + z = 0 \) → \( 1x - 2y + 1z = 0 \)
2. \( 0x + 1y - 1z = 2 \) → \( 0x + 1y - 1z = 2 \)
3. \( 2x + 0y - 3z = 10 \) → \( 2x + 0y - 3z = 10 \)

### Step 2: Form the coefficient matrix \( A \), variable matrix \( X \), and constant matrix \( B \)
The coefficient matrix \( A \), variable matrix \( X \), and constant matrix \( B \) can be defined as follows:
\[ A = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 2 & 0 & -3 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 0 \\ 2 \\ 10 \end{pmatrix} \]

### Step 3: Find the determinant of matrix \( A \)
To find the determinant of matrix \( A \):
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ 0 & -3 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 0 & -1 \\ 2 & -3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 1 \\ 2 & 0 \end{vmatrix} \]
Calculating the determinants:
\[ \text{det}(A) = 1 \cdot (1 \cdot -3 - 0 \cdot -1) + 2 \cdot (0 \cdot -3 - 2 \cdot -1) + 1 \cdot (0 \cdot 0 - 2 \cdot 1) \]
\[ = -3 + 2 \cdot 2 + 0 = -3 + 4 = 1 \]

### Step 4: Find the inverse of matrix \( A \)
Since the determinant is non-zero, we can find the inverse of matrix \( A \):
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adj}(A) \]
To find the adjoint \( \text{Adj}(A) \), we need to calculate the cofactor matrix and then take its transpose.

### Step 5: Calculate the adjoint of matrix \( A \)
The cofactor matrix is calculated as follows:
\[ \text{Cofactor}(A) = \begin{pmatrix} \begin{vmatrix} 1 & -1 \\ 0 & -3 \end{vmatrix} & -\begin{vmatrix} 0 & -1 \\ 2 & -3 \end{vmatrix} & \begin{vmatrix} 0 & 1 \\ 2 & 0 \end{vmatrix} \\ -\begin{vmatrix} -2 & 1 \\ 0 & -3 \end{vmatrix} & \begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix} & -\begin{vmatrix} 1 & -2 \\ 2 & 0 \end{vmatrix} \\ \begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} & -\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} & \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} \end{pmatrix} \]
Calculating each of these determinants will give us the cofactor matrix, and taking the transpose will yield the adjoint matrix.

### Step 6: Multiply the inverse of \( A \) with \( B \)
Once we have \( A^{-1} \), we can find the solution by multiplying \( A^{-1} \) with \( B \):
\[ X = A^{-1}B \]

### Step 7: Solve for \( x, y, z \)
The resulting matrix \( X \) will give us the values of \( x, y, z \).