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ECIV 301
Programming & Graphics
Numerical Methods for Engineers
Lecture 17
Solution of Systems of Equations
Last Time Linear Equations in Matrix Form
10z8y3x5
6z3yx12
24z23y6x10
23610
3112
835
z
y
x
24
6
10 10
z
y
x
835
6
z
y
x
3112
24
z
y
x
23610
# Equations = # Unknowns = n
Square Matrix n x n
Last Time Solution of Linear Equations
10
7
9
500
310
835
z
y
x
Express In Matrix Form
Upper Triangular
What is the characteristic?
Solution by Back Substitution
Last Time Solution of Linear Equations
Objective
Can we express any system of equations in a form
nnnn
n
n
n
b
b
b
b
x
x
x
x
a
aa
aaa
aaaa
3
2
1
3
2
1
333
22322
1131211
000
00
0
0
Last Time Background
Consider
1035 yx(Eq 1)
5810 yx(Eq 2)
Solution
5.7
5.6
y
x
20610 yx2*(Eq 1)
5810 yx(Eq 2)
Solution
5.7
5.6
y
x!!!!!!
Scaling Does Not Change the SolutionScaling Does Not Change the Solution
Last Time Background
Consider
20610 yx(Eq 1)
152 y(Eq 2)-(Eq 1)
Solution
5.7
5.6
y
x!!!!!!
20610 yx(Eq 1)
5810 yx(Eq 2)
Solution
5.7
5.6
y
x
Operations Do Not Change the SolutionOperations Do Not Change the Solution
Last Time Gauss Elimination
10835 zyx
2423610 zyx
6312 zyx
Example
Forward Elimination
Last Time Gauss Elimination
Forward Elimination
064.62
30
10
645.700
2.162.60
835
z
y
x
Last TimeGauss Elimination
Back Substitution
118.8645.7/064.62 z
0502.26
2.6
118.82.1630
y
6413.0
5
118.880502.26310
x
064.62
30
10
645.700
2.162.60
835
z
y
x
Last Time GE – Potential Problem
10830 zyx
2423610 zyx
6312 zyx
Forward Elimination
Gauss Elimination – Potential Problem
10830 zyx
2423610 zyx
6312 zyx
0
12 Division By Zero!!Operation Failed
Gauss Elimination – Potential Problem
10830 zyx
2423610 zyx
6312 zyx
12
0OK!!
Gauss Elimination – Potential Problem
10830 zyx
2423610 zyx
6312 zyx
Pivoting
6312 zyx
10830 zyx
Partial Pivoting
nn
nnnnn
lll
n
n
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaaaaaa
aaaa
3
2
1
3
2
1
321
ln321
3333231
2232221
1131211
a32>a22
al2>a22
NO
YES
Partial Pivoting
nn
nnnnn
n
n
lll
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaaaaaa
aaaa
3
2
1
3
2
1
321
2232221
3333231
ln321
1131211
Full Pivoting
• In addition to row swaping
• Search columns for max elements
• Swap Columns
• Change the order of xi
• Most cases not necessary
EXAMPLE
4.71
3.19
85.7
102.03.0
3.071.0
2.01.03
3
2
1
x
x
x
Eliminate Column 1
3
1.0
PIVOTS
4.71
3.19
85.7
102.03.0
3.071.0
2.01.03
3
3.0
Eliminate Column 1
6150.70
5617.19
85.7
0200.1019000.00
29333.000333.70
2.01.03
Eliminate Column 2
00333.7
19000.0
PIVOTS
6150.70
5617.19
85.7
0200.1019000.00
29333.000333.70
2.01.03
Eliminate Column 2
0843.70
5617.19
85.7
01200.1000
29333.000333.70
2.01.03
LU DecompositionPIVOTS
Column 1PIVOTS
Column 2
03333.0
1.0 02713.0
LU Decomposition
As many as, and in the location of, zeros
UpperTriangular
MatrixU
01200.1000
29333.000333.70
2.01.03
LU DecompositionPIVOTS
Column 1
PIVOTSColumn 2
LowerTriangular
Matrix
1
1
1
0
0
0
L
03333.0
1.0 02713.0
LU Decomposition
102713.01.0
0103333.0
001
=
This is the original matrix!!!!!!!!!!
01200.1000
29333.000333.70
2.01.03
102.03.0
3.071.0
2.01.03
LU Decomposition
4.71
3.19
85.7
102713.01.0
0103333.0
001
3
2
1
y
y
y
4.71
3.19
85.7
102.03.0
3.071.0
2.01.03
3
2
1
x
x
x
L y b
LU Decomposition
4.71
3.19
85.7
102713.01.0
0103333.0
001
3
2
1
y
y
y
L y b
85.71 y
5617.190333.03.19 12 yy
0843.70)02713.0(1.04.71 213 yyy
LU Decomposition85.71 y
5617.190333.03.19 12 yy
0843.70)02713.0(1.04.71 213 yyy
0843.70
5617.19
85.7
01200.1000
29333.000333.70
2.01.03
LU Decomposition
• Ax=b
• A=LU - LU Decomposition
• Ly=b- Solve for y
• Ux=y - Solve for x
