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1. Use LU-factorization method (discussed in the handout) to solve the system b 2 1 40 0-10 3 -1 1 1 3 0 0 2 A 1 20-2 18 x1 X2 X3 CA X5 = =

1. Use LU-factorization method (discussed in the handout) to solve the system b 2 1 40 0-10 3 -1 1 1 3 0 0 2 A 1 20-2 18 x1 X2 X3 CA X5 = =

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Problem 1. Use LU-factorization method (discussed in the handout) to solve the system 6 A te I = 2 1 40 1 X2 0-10 3 -1 60-0 0-1 1 3 2 1 2 0 -2 =
Transcribed Image Text:Problem 1. Use LU-factorization method (discussed in the handout) to solve the system 6 A te I = 2 1 40 1 X2 0-10 3 -1 60-0 0-1 1 3 2 1 2 0 -2 =
Given a matrix A. Suppose we can write A as the product A = LU where L is a (square) lower triangular matrix and U is a row echelon matrix. This product is called the "LU- factorization of A". An application of LU-Factorization Suppose we want to solve the system Ar= b. If A = LU then Az = (LU) L(Ux) =b Now if we let then A= Ly and Ly = b Now in order to solve the original system A = 6, we solve the system (2) first and then the system (1). ● Algorithm: Solving systems by LU-factorization Consider the system A = b where A is an m x n matrix. In this section we will learn an algorithm to find the matrices L and U for A. Step 1, write the augmented matrix [A | Im] Step 2, transform A into its echelon form. We get [UL-¹] Solution. • Step 3, solve Ly= b or equivalently = L-¹6, and then solve y = UT. Example 2. Using LU-factorization, solve Step 1, write Step 2 Row echelon: Therefore, U = y=Uz Y1 Step 3 Let y = y2 Y3 = A 201 22 1 4 22 1 5, 2012 0 2 0 2 and L-¹ = 0001 Now we solve y = U, we get x1 = x1 X2 X3 X4 2012 1 0 0 221 4 01 0 2 2 1 500 1 Clearly, x3 = t is a free variable and -t + 5 = and solve y L-¹6, we get = 2012 1 0 0 0202-1 1 0 0001 0 -1 1 1 0 2 [201 0202 0001 1 D-60-Q -1 -1 1 6 (-) 5 3 -1 1 11-0 4 -, x₂ = 4, 4 = -2 2
Transcribed Image Text:Given a matrix A. Suppose we can write A as the product A = LU where L is a (square) lower triangular matrix and U is a row echelon matrix. This product is called the "LU- factorization of A". An application of LU-Factorization Suppose we want to solve the system Ar= b. If A = LU then Az = (LU) L(Ux) =b Now if we let then A= Ly and Ly = b Now in order to solve the original system A = 6, we solve the system (2) first and then the system (1). ● Algorithm: Solving systems by LU-factorization Consider the system A = b where A is an m x n matrix. In this section we will learn an algorithm to find the matrices L and U for A. Step 1, write the augmented matrix [A | Im] Step 2, transform A into its echelon form. We get [UL-¹] Solution. • Step 3, solve Ly= b or equivalently = L-¹6, and then solve y = UT. Example 2. Using LU-factorization, solve Step 1, write Step 2 Row echelon: Therefore, U = y=Uz Y1 Step 3 Let y = y2 Y3 = A 201 22 1 4 22 1 5, 2012 0 2 0 2 and L-¹ = 0001 Now we solve y = U, we get x1 = x1 X2 X3 X4 2012 1 0 0 221 4 01 0 2 2 1 500 1 Clearly, x3 = t is a free variable and -t + 5 = and solve y L-¹6, we get = 2012 1 0 0 0202-1 1 0 0001 0 -1 1 1 0 2 [201 0202 0001 1 D-60-Q -1 -1 1 6 (-) 5 3 -1 1 11-0 4 -, x₂ = 4, 4 = -2 2
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