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Let's solve the system of linear equations using LU decomposition. The system is:

\[
\begin{cases}
3x + 2y - z = 1 \\
x + 2y + 2z = 0 \\
2x + y + 3z = -1
\end{cases}
\]

### Step 1: Write the system in matrix form

The matrix equation for the system \( AX = B \) is:

\[
A = \begin{bmatrix} 3 & 2 & -1 \\ 1 & 2 & 2 \\ 2 & 1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}.
\]

### Step 2: Perform LU decomposition

We will decompose matrix \( A \) into the product of a lower triangular matrix \( L \) and an upper triangular matrix \( U \), i.e., \( A = LU \). Let's go through the process step-by-step.

1. **Initialize \( L \) and \( U \):**
   
   \[
   L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}.
   \]

2. **Compute \( U \) and \( L \):**
   - For \( U_{11} = 3 \), \( U_{12} = 2 \), \( U_{13} = -1 \).
   - For \( L_{21} = \frac{1}{3} \), update \( U \) and \( L \):
     \[
     U_{22} = 2 - \left(\frac{1}{3} \times 2\right) = \frac{4}{3}, \quad U_{23} = 2 - \left(\frac{1}{3} \times (-1)\right) = \frac{7}{3}.
     \]
   - For \( L_{31} = \frac{2}{3} \), update \( U \):
     \[
     U_{32} = 1 - \left(\frac{2}{3} \times 2\right) = -\frac{1}{3}, \quad U_{33} = 3 - \left(\frac{2}{3} \times (-1)\right) = \frac{11}{3}.
     \]
   - For \( L_{32} = \frac{-1}{\frac{4}{3}} = -\frac{3}{4} \).

3. **Final matrices:**
   \[
   L = \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{3} & 1 & 0 \\ \frac{2}{3} & -\frac{3}{4} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} 3 & 2 & -1 \\ 0 & \frac{4}{3} & \frac{7}{3} \\ 0 & 0 & \frac{11}{3} \end{bmatrix}.
   \]

### Step 3: Solve the system \( LUx = B \)

1. **Solve \( Ly = B \):**
   
   \[
   \begin{aligned}
   y_1 &= 1, \\
   y_2 &= 0 - \frac{1}{3} \times 1 = -\frac{1}{3}, \\
   y_3 &= -1 - \left(\frac{2}{3} \times 1\right) + \left(-\frac{3}{4} \times -\frac{1}{3}\right) = -\frac{11}{12}.
   \end{aligned}
   \]

2. **Solve \( Ux = y \):**
   
   \[
   \begin{aligned}
   x_3 &= \frac{y_3}{U_{33}} = \frac{-\frac{11}{12}}{\frac{11}{3}} = -\frac{1}{4}, \\
   x_2 &= \frac{y_2 - U_{23}x_3}{U_{22}} = \frac{-\frac{1}{3} - \frac{7}{3} \times \left(-\frac{1}{4}\right)}{\frac{4}{3}} = 0, \\
   x_1 &= \frac{y_1 - U_{12}x_2 - U_{13}x_3}{U_{11}} = \frac{1 - 2 \times 0 - (-1) \times \left(-\frac{1}{4}\right)}{3} = \frac{3}{4}.
   \end{aligned}
   \]

### Solution

The solution to the system is:

\[
x = \frac{3}{4}, \quad y = 0, \quad z = -\frac{1}{4}.
\]

Would you like further details on any of these steps, or have more questions?

### 5 Related Questions
1. How can LU decomposition be used to find the determinant of a matrix?
2. What are the advantages of using LU decomposition over other methods for solving linear systems?
3. How do you handle pivoting in LU decomposition for matrices that are not diagonally dominant?
4. Can LU decomposition be applied to non-square matrices?
5. What is the relationship between LU decomposition and Gaussian elimination?

### Tip
LU decomposition is particularly useful for solving multiple systems of equations with the same coefficient matrix, as it allows you to decompose the matrix once and reuse it to find solutions for different right-hand sides.

Learn how to solve a system of linear equations using LU decomposition. This detailed solution breaks down the steps of decomposing a matrix into lower and upper triangular forms and solving the system using forward and backward substitution. Ideal for undergraduate students studying linear algebra.

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