Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 3DE
Related questions
Question
please answer question 1,this is an example ,you can make a reference thanks !
![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
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F655e5d67-ab19-404c-b883-3aa3c693f6a1%2Fe10a32ed-c11a-451c-ac89-b6acbb1841a0%2Ff1hv9lc_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F655e5d67-ab19-404c-b883-3aa3c693f6a1%2Fe10a32ed-c11a-451c-ac89-b6acbb1841a0%2F2gdy0sj_processed.jpeg&w=3840&q=75)
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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