Ch8_2
8.1 Gaussian EliminationDefinitionA matrix is in echelon form if
1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix.
2. The first nonzero element of each row is 1. This element is called a leading 1.
3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row.
(This implies that all the elements below a leading 1 are zero.)
Reduced echelon form 與 echelon from 的差異: echelon form 的 leading 1 上面的數字不必為零
Ch8_3
Example 1Solving the following system of linear equations using the method of Gaussian elimination.
Solution 412842222
1232
4321
4321
4321
xxxxxxxxxxxx
Starting with the augmented matrix, create zeros below the pivot in the first column.
4128422122112321
682001310012321
1)2(312RR
RR
At this stage, we create a zero only below the pivot.
420001310012321
2)2(3 RR
210001310012321
32
1R
Echelon formWe have arrived at the echelon form.
Ch8_4
The corresponding system of equation is
213
1232
4
43
4321
xxxxxxx
We get
51)2(3
3
3
xx
Substituting x4 = 2 and x3 = 5 into the first equation,
102102
1)2(2)5(32
21
21
21
xxxx
xx
Let x2 = r. The system has many solutions. The solutions are2 ,5 , ,102 4321 xxrxrx
Ch8_5
Example 2Solving the following system of linear equations using the method of Gaussian elimination, performing back substitution using matrices.
412842222
1232
4321
4321
4321
xxxxxxxxxxxx
SolutionWe arrive at the echelon form as in the previous example.
4128422122112321
210001310012321
Echelon formThis marks the end of the forward elimination of variables from equations. We now commence the back substitution using matrices.
Ch8_6
210005010050321
3)3(2
)2(1 210001310012321
RRRR 2
2100050100
100021 )3(1 RR 3
This matrix is the reduced echelon form of the original augmented matrix. The corresponding system of equations is
25
102
4
3
21
xxxx
Let x2 = r. We get same solution as previously,2 ,5 , ,102 4321 xxrxrx
Ch8_7
Comparison of Gauss-Jordan and Gaussian Elimination
Count of Operations for n n system with Unique Solution
Number of Multiplications
Number of Additions
Gauss-Jordan
Gaussian elimination
) largefor (
323
222n
nnn 222
33 nnn
333
32
3 nnn
n 36
523
323 nnnn
Ch8_8
8.2 The Method of LU Decomposition
8204
0725
0013
0002
7000
2400
1920
3528
lower triangular matrix upper triangular matrix
Ax=y A=LU …
Ch8_9
Example 1Solving the following system of equations, which has a triangular matrix of coefficients.
354
2 3
8 2
321
21
1
xxx
xx
x
Solution
1st equations gives 482 11 xx
2nd equation gives 223423 2221 xxxx
3rd equation gives 331016354 33321 xxxxx
The solution is .3,2,4 321 xxx
By forward substitution:
Ch8_10
Example 2Solving the following system of equations, which has an upper triangular matrix of coefficients.
63
4
042
3
32
321
x
xx
xxx
Solution
3rd equation gives 263 33 xx
2nd equation gives 2424 2232 xxxx
1st equations gives 50822042 11321 xxxxx
The solution is .2,2,5 321 xxx
By back substitution:
Ch8_11
DefinitionLet A be a square matrix that can be factored into the form A = LU, where L is a lower triangular matrix and U is an upper triangular matrix, This factoring is called an LU decomposition of A. (Not every matrix has an LU decomposition, and when it exists, it is not unique.)
Ch8_12
Solution of AX = B1. Find the LU decomposition of A.
(If A has no LU decomposition, the method is not applicable.)
2. Solve LY = B by forward substitution.
3. Solve UX = Y by back substitution.
Let AX = B be a system of n equations in n variables, where A has LU decomposition A = LU. LUX = B two subsystems: UX = Y (upper triangular) and LY = B (lower triangular)
Method of LU Decomposition
How to decompose A=LU? 利用 elementary matrices
Ch8_13
Elementary Matrices (複習第二章 )
Definition An elementary matrix is one that can be obtained from the identity matrix In through a single elementary row operation.
Example
100
010
001
3I
010
100
001
1ER2 R3
100
050
001
2E5R2
100
012
001
3ER2+ 2R1
Ch8_14
Elementary Matrices (複習第二章 )
ihg
fed
cba
A
AEA
fed
ihg
cba
1
010
100
001
R2 R3
AEA
ihg
fed
cba
2
100
050
001
555
5R2
AEA
ihg
cfbead
cba
3
100
012
001
222
R2+ 2R1
一個矩陣做 elementary row operation ,相當於在左邊乘一個對應的 elementary matrix 。
Ch8_15
Elementary matrices (複習第二章 )Each elementary matrix is invertible.
Example1
221EI
RR
100
010
021
1E
100
010
001
I 112
100
010
021
EE
, 221
1 IERR
IEE 12 i.e.,
Note that (E1)12 = (E2)12.
Ch8_16
( 利用 elementary matrix)A … U (upper triangular) U = Ek E1 A A = (E1)1 (Ek)1 U
If each such elementary matrix Ei is a lower triangular matrices,it can be proved that (E1)1, , (Ek)1 are lower triangular, and(E1)1 (Ek)1 is a lower triangular matrix.Let L=(E1)1 (Ek)1 then A=LU.
How to decompose A=LU?
Ch8_17
Example 3Solving the following system of equations using LU decomposition.
21226574
132
321
321
321
xxxxxxxxx
SolutionLet us transform the matrix of coefficients A into upper triangular form U by creating zeros below the main diagonal as follows.
1226714312
A
1226110312
122 RR
310110312
133 RR
200110312
23 RR
U
Ch8_18
The elementary matrices that correspond to these row operations are
11001000123
,
103010001133
,
100012001122
321
RR
E
RR
E
RR
E
The inverse of these matrices are
110010001
,103010001
,100012001
13
12
11 EEE
We get
113012001
13
12
11 EEEL
Thus
200110312
113012001
UL
A
Ch8_19
We now solve the given system LUX = B by solving the two subsystems LY = B and UX = Y. We get
2
5
1
113
012
001
:
3
2
1
y
y
y
BLY
This lower triangular system has solution .2,7,1 321 yyy
271
200110312
:
3
2
1
xxx
YUX
The upper triangular system has solution .1,8,5 321 xxx
The solution to the given system is .1,8,5 321 xxx
Ch8_20
Construction of a LU decomposition of a Matrix A
1. Use row operations to arrive at U.
(The operations must involve adding multiples of rows to rows. In general, if row interchanges are required to arrive at U, an LU form does not exists.)
2. The diagonal element of L are ls.
The nonzero elements of L corresponding to row operations.
The row operation Ri + cRj implies that lij = c.
Ch8_21
Example 4Solve the following system if equations using LU decomposition.
5823841235
1243
321
321
321
xxxxxxxxx
Solution
2384
351
431
740
120
431
143
12
RR
RR
500
120
431
223 RR
These row operations lead to the following LU decomposition of A.
500
120
431
124
011
001
AL21= 1
L31= L32=
Ch8_22
We again solve the given system LUX = B by solving the two subsystems LY = B and UX = Y.
58
12
12
124
011
001
:
3
2
1
y
y
y
BLY
10
0
12
500
120
431
:
3
2
1
x
x
x
YUX
This lower triangular system has solution y1 = 12, y2 = 0, y3 = 10.
This upper triangular system has solution x1 = 1, x2 = 1, x3 = 2.
The solution to the given system is x1 = 1, x2 = 1, x3 = 2.