Math Problem Statement

A =

2 −8 0

3 −7 8 −1 4 2

1 0 0

3 2 1 0 − 1 2 0 1

2 −8 0

0 5 8 0 0 2 , b =

10

−1 −9 Solve the system Ax = b using the given LU factorization of A by first solving Ly = b for y

Solution

Given the LU factorization of matrix AA, we need to solve the system Ax=bAx = b. The LU factorization expresses AA as the product of a lower triangular matrix LL and an upper triangular matrix UU, i.e., A=LUA = LU.

The process involves two steps:

  1. Solve the system Ly=bLy = b for yy (forward substitution).
  2. Solve the system Ux=yUx = y for xx (backward substitution).

We are given:

  • The matrix AA,
2 & -8 & 0 \\ 3 & -7 & 8 \\ -1 & 4 & 2 \end{pmatrix}$$ - The LU factorization, $$L = \begin{pmatrix} 1 & 0 & 0 \\ \frac{3}{2} & 1 & 0 \\ -\frac{1}{2} & \frac{1}{2} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} 2 & -8 & 0 \\ 0 & 5 & 8 \\ 0 & 0 & 2 \end{pmatrix}$$ - The vector $$b = \begin{pmatrix} 10 \\ -1 \\ -9 \end{pmatrix}$$. ### Step 1: Solve $$Ly = b$$ We need to solve: $$L \cdot y = b$$ Where $$y = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}$$ and $$L$$ is the lower triangular matrix: $$\begin{pmatrix} 1 & 0 & 0 \\ \frac{3}{2} & 1 & 0 \\ -\frac{1}{2} & \frac{1}{2} & 1 \end{pmatrix} \cdot \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 10 \\ -1 \\ -9 \end{pmatrix}$$ This leads to the following system of equations: 1. $$y_1 = 10$$ 2. $$\frac{3}{2} y_1 + y_2 = -1$$ 3. $$-\frac{1}{2} y_1 + \frac{1}{2} y_2 + y_3 = -9$$ #### Solve for $$y_1$$, $$y_2$$, and $$y_3$$: 1. $$y_1 = 10$$ 2. $$\frac{3}{2} (10) + y_2 = -1$$ $$15 + y_2 = -1$$ $$y_2 = -1 - 15 = -16$$ 3. $$-\frac{1}{2} (10) + \frac{1}{2} (-16) + y_3 = -9$$ $$-5 - 8 + y_3 = -9$$ $$y_3 = -9 + 13 = 4$$ Thus, we have: $$y = \begin{pmatrix} 10 \\ -16 \\ 4 \end{pmatrix}$$ ### Step 2: Solve $$Ux = y$$ Next, we solve: $$U \cdot x = y$$ Where $$x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$ and $$U$$ is the upper triangular matrix: $$\begin{pmatrix} 2 & -8 & 0 \\ 0 & 5 & 8 \\ 0 & 0 & 2 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 10 \\ -16 \\ 4 \end{pmatrix}$$ This leads to the following system of equations: 1. $$2x_1 - 8x_2 = 10$$ 2. $$5x_2 + 8x_3 = -16$$ 3. $$2x_3 = 4$$ #### Solve for $$x_1$$, $$x_2$$, and $$x_3$$: 1. From equation (3): $$2x_3 = 4$$ $$x_3 = 2$$ 2. From equation (2): $$5x_2 + 8(2) = -16$$ $$5x_2 + 16 = -16$$ $$5x_2 = -32$$ $$x_2 = -\frac{32}{5} = -6.4$$ 3. From equation (1): $$2x_1 - 8(-6.4) = 10$$ $$2x_1 + 51.2 = 10$$ $$2x_1 = 10 - 51.2 = -41.2$$ $$x_1 = \frac{-41.2}{2} = -20.6$$ Thus, the solution is: $$x = \begin{pmatrix} -20.6 \\ -6.4 \\ 2 \end{pmatrix}$$ ### Final Answer: The solution to the system $$Ax = b$$ is: $$x = \begin{pmatrix} -20.6 \\ -6.4 \\ 2 \end{pmatrix}$$ --- Would you like more details on any step or further clarification? Here are some related questions: 1. What are the advantages of LU factorization in solving systems of equations? 2. Can LU factorization be applied to any matrix? 3. How does forward substitution work in triangular systems? 4. What are some efficient numerical methods for solving large linear systems? 5. How do you determine the LU factorization of a given matrix? **Tip:** LU factorization is particularly useful for solving multiple systems with the same matrix but different right-hand sides, as the factorization only needs to be done once.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
LU Factorization
Systems of Linear Equations
Forward and Backward Substitution

Formulas

LU factorization: A = L * U
Forward substitution: L * y = b
Backward substitution: U * x = y

Theorems

LU Factorization Theorem

Suitable Grade Level

Undergraduate Level (Linear Algebra)