Math Problem Statement

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

Solution

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

  1. Matrix AA is factorized into two matrices: LL (lower triangular) and UU (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 YY in the equation LY=bLY = b: LY=bL \cdot Y = b We solve the lower triangular system, and the solution is:

    -2 \\ -1 \\ 2 \\ -3 \end{pmatrix}$$
  3. Solving for XX in the equation UX=YUX = Y: UX=YU \cdot X = Y We solve the upper triangular system, and the solution to the system AX=bAX = b is:

    -2 \\ -1 \\ 2 \\ -3 \end{pmatrix}$$

Thus, the solution to the system is X=(2123)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 AA 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 AA but different vectors bb.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Factorization

Formulas

LU Decomposition: A = LU
Solving LY = b
Solving UX = Y

Theorems

LU Factorization

Suitable Grade Level

Undergraduate Level