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Here is the step-by-step solution to the system \( AX = b \) using LU factorization:

1. **Matrix \( A \)** is factorized into two matrices: \( L \) (lower triangular) and \( U \) (upper triangular) using LU decomposition:
   - \( L = \begin{pmatrix}
       1 & 0 & 0 & 0 \\
       2 & 1 & 0 & 0 \\
       -1 & 0 & 1 & 0 \\
       -4 & 3 & -5 & 1
     \end{pmatrix} \)
   - \( U = \begin{pmatrix}
       1 & -2 & -4 & -3 \\
       0 & -3 & 1 & 0 \\
       0 & 0 & 2 & 1 \\
       0 & 0 & 0 & 1
     \end{pmatrix} \)

2. **Solving for** \( Y \) in the equation \( LY = b \):
   \[
   L \cdot Y = b
   \]
   We solve the lower triangular system, and the solution is:
   \[
   Y = \begin{pmatrix}
     -2 \\
     -1 \\
      2 \\
     -3
   \end{pmatrix}
   \]

3. **Solving for** \( X \) in the equation \( UX = Y \):
   \[
   U \cdot X = Y
   \]
   We solve the upper triangular system, and the solution to the system \( AX = b \) is:
   \[
   X = \begin{pmatrix}
     -2 \\
     -1 \\
      2 \\
     -3
   \end{pmatrix}
   \]

Thus, the solution to the system is \( X = \begin{pmatrix} -2 \\ -1 \\ 2 \\ -3 \end{pmatrix} \).

Would you like more details or clarifications on any step?

Here are some related questions to explore:
1. How does LU factorization work in general?
2. What are the differences between LU and QR factorization?
3. Can LU factorization always be applied to any matrix?
4. How would the solution change if \( A \) were singular?
5. What are the practical applications of LU factorization?

**Tip:** LU factorization is particularly useful when solving multiple systems with the same matrix \( A \) but different vectors \( b \).

Find the solution of the system AX = b using LU factorization, where A and b are given matrices.

Learn how to solve the system AX = b using LU factorization, where A is a 4x4 matrix and b is a vector. This detailed step-by-step solution covers key concepts from linear algebra, including matrix factorization and solving triangular systems. Suitable for undergraduate students.

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