Math Problem Statement

Use the given LU factorization to find all solutions to Ax = b, where A = [[1, 0, 0, 0], [-4, 1, 0, 0], [0, -5, 1, 0], [9, -3, 7, 1]] and b = [[65], [-270], [62], [699]]

Solution

The given matrix equation Ax=bA \mathbf{x} = \mathbf{b} requires solving using LU factorization, where matrix AA has already been factorized into matrices LL (lower triangular) and UU (upper triangular):

A=LUA = L U

We can rewrite the system as two steps:

  1. Solve Ly=bL \mathbf{y} = \mathbf{b} for y\mathbf{y} (forward substitution).
  2. Solve Ux=yU \mathbf{x} = \mathbf{y} for x\mathbf{x} (back substitution).

Step 1: Solve Ly=bL \mathbf{y} = \mathbf{b}

The lower triangular matrix LL is derived from the given matrix structure. From visual inspection, we have:

1 & 0 & 0 & 0 \\ -4 & 1 & 0 & 0 \\ 0 & -5 & 1 & 0 \\ 9 & -3 & 7 & 1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 65 \\ -270 \\ 62 \\ 699 \end{pmatrix}$$ Now, perform forward substitution to solve for $$\mathbf{y}$$. $$L \mathbf{y} = \mathbf{b} \quad \Rightarrow \begin{aligned} & y_1 = 65 \\ & -4 y_1 + y_2 = -270 \quad \Rightarrow y_2 = -10 \\ & -5 y_2 + y_3 = 62 \quad \Rightarrow y_3 = 12 \\ & 9 y_1 - 3 y_2 + 7 y_3 + y_4 = 699 \quad \Rightarrow y_4 = -200 \end{aligned}$$ Thus, $$\mathbf{y} = \begin{pmatrix} 65 \\ -10 \\ 12 \\ -200 \end{pmatrix}$$. ### Step 2: Solve $$U \mathbf{x} = \mathbf{y}$$ The upper triangular matrix $$U$$ is: $$U = \begin{pmatrix} 5 & -5 & -5 & -5 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ Now, perform back substitution to solve for $$\mathbf{x}$$. $$U \mathbf{x} = \mathbf{y} \quad \Rightarrow \begin{aligned} & 4 x_4 = 12 \quad \Rightarrow x_4 = 3 \\ & -x_3 + x_4 = -10 \quad \Rightarrow x_3 = 13 \\ & 5x_1 - 5x_2 - 5x_3 - 5x_4 = 65 \quad \Rightarrow x_1 - x_2 - x_3 - x_4 = 13 \quad \Rightarrow \text{no unique solution}. \end{aligned}$$ The system is underdetermined, so there is a dependency in the solution. The general solution involves free variables. Let me know if you would like further explanation or details on a specific part of the process! Also, here are 5 related questions to expand this concept: 1. How does LU factorization help in solving linear systems? 2. What are the conditions for the existence of a unique solution in linear systems? 3. How do we handle systems with infinitely many solutions in linear algebra? 4. What other matrix decomposition methods are commonly used in solving $$A \mathbf{x} = \mathbf{b}$$? 5. Can LU factorization always be performed on any square matrix? **Tip:** LU factorization is particularly efficient for solving systems of equations when the same matrix $$A$$ is used with multiple right-hand sides (i.e., multiple $$\mathbf{b}$$ vectors).

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
LU Factorization
Systems of Linear Equations

Formulas

Ax = b
LU decomposition: A = LU
Forward substitution for Ly = b
Back substitution for Ux = y

Theorems

LU Factorization Theorem
Existence and Uniqueness Theorem for Linear Systems

Suitable Grade Level

Undergraduate (College Level)