gaussian elimination
DESCRIPTION
Gaussian Elimination. Major: All Engineering Majors Author(s): Autar Kaw http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates. Naïve Gauss Elimination. Naïve Gaussian Elimination. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/1.jpg)
Gaussian Elimination
Major: All Engineering Majors
Author(s): Autar Kaw
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
![Page 2: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/2.jpg)
Naïve Gauss Elimination
![Page 3: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/3.jpg)
Naïve Gaussian EliminationA method to solve simultaneous linear equations of the form [A][X]=[C]Two steps1. Forward Elimination2. Back Substitution
![Page 4: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/4.jpg)
Forward Elimination
2.2792.1778.106
11214418641525
3
2
1
xxx
The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix
735.021.968.106
7.00056.18.40
1525
3
2
1
xxx
![Page 5: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/5.jpg)
Forward EliminationA set of n equations and n unknowns
11313212111 ... bxaxaxaxa nn
22323222121 ... bxaxaxaxa nn
nnnnnnn bxaxaxaxa ...332211
. . . . . .
(n-1) steps of forward elimination
![Page 6: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/6.jpg)
Forward EliminationStep 1 For Equation 2, divide Equation 1 by and multiply by .
)...( 1131321211111
21 bxaxaxaxaaa
nn
111
211
11
21212
11
21121 ... b
aaxa
aaxa
aaxa nn
11a21a
![Page 7: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/7.jpg)
Forward Elimination
111
211
11
21212
11
21121 ... b
aaxa
aaxa
aaxa nn
111
2121
11
212212
11
2122 ... b
aabxa
aaaxa
aaa nnn
'2
'22
'22 ... bxaxa nn
22323222121 ... bxaxaxaxa nn Subtract the result from Equation 2.
−_________________________________________________
or
![Page 8: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/8.jpg)
Forward EliminationRepeat this procedure for the remaining equations to reduce the set of equations as
11313212111 ... bxaxaxaxa nn '2
'23
'232
'22 ... bxaxaxa nn
'3
'33
'332
'32 ... bxaxaxa nn
''3
'32
'2 ... nnnnnn bxaxaxa
. . . . . . . . .
End of Step 1
![Page 9: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/9.jpg)
Step 2Repeat the same procedure for the 3rd term of Equation 3.
Forward Elimination
11313212111 ... bxaxaxaxa nn
'2
'23
'232
'22 ... bxaxaxa nn
"3
"33
"33 ... bxaxa nn
""3
"3 ... nnnnn bxaxa
. . . . . .
End of Step 2
![Page 10: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/10.jpg)
Forward EliminationAt the end of (n-1) Forward Elimination steps, the system of equations will look like
'2
'23
'232
'22 ... bxaxaxa nn
"3
"33
"33 ... bxaxa nn
11 nnn
nnn bxa
. . . . . .
11313212111 ... bxaxaxaxa nn
End of Step (n-1)
![Page 11: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/11.jpg)
Matrix Form at End of Forward Elimination
)(n-n
"
'
n)(n
nn
"n
"
'n
''n
b
bbb
x
xxx
a
aaaaaaaaa
1
3
2
1
3
2
1
1
333
22322
1131211
0000
000
![Page 12: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/12.jpg)
Back SubstitutionSolve each equation starting from the last equation
Example of a system of 3 equations
735.021.968.106
7.00056.18.40
1525
3
2
1
xxx
![Page 13: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/13.jpg)
Back Substitution Starting Eqns
'2
'23
'232
'22 ... bxaxaxa nn
"3
"3
"33 ... bxaxa nn
11 nnn
nnn bxa
. .
. . . .
11313212111 ... bxaxaxaxa nn
![Page 14: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/14.jpg)
Back SubstitutionStart with the last equation because it has only one unknown
)1(
)1(
nnn
nn
n ab
x
![Page 15: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/15.jpg)
Back Substitution
1,...,1for11
11
nia
xabx i
ii
n
ijj
iij
ii
i
)1(
)1(
nnn
nn
n ab
x
1,...,1for...
1
1,2
12,1
11,
1
nia
xaxaxabx i
ii
ninii
iiii
iii
ii
i
![Page 16: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/16.jpg)
THE END
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Example 1The upward velocity of a rocket is given at three different times
Time, Velocity, 5 106.88 177.212 279.2
The velocity data is approximated by a polynomial as:
12.t5 , 322
1 atatatv
Find the velocity at t=6 seconds .
s t m/s vTable 1 Velocity vs. time data.
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Example 1 Cont. Assume
12.t5 ,atatatv 322
1
3
2
1
323
222
121
111
vvv
aaa
tttttt
3
2
1
Results in a matrix template of the form:
Using data from Table 1, the matrix becomes:
2.2792.1778.106
11214418641525
3
2
1
aaa
![Page 19: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/19.jpg)
Example 1 Cont.
22791121442177186481061525
227921778106
11214418641525
3
2
1
.
.
.
.
.
.
aaa
1. Forward Elimination2. Back Substitution
![Page 20: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/20.jpg)
Forward Elimination
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Number of Steps of Forward Elimination
Number of steps of forward elimination is
(n1)(31)2
![Page 22: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/22.jpg)
Divide Equation 1 by 25 andmultiply it by 64, .
Forward Elimination: Step 1
.
408.27356.28.126456.28.1061525
208.96 56.18.4 0
408.273 56.2 8.1264177.2 1 8 64
2.2791121442.17718648.1061525
2.279112144208.9656.18.408.1061525
56.22564
Subtract the result from Equation 2
Substitute new equation for Equation 2
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Forward Elimination: Step 1 (cont.)
.
168.61576.58.2814476.58.1061525
2.279112144208.9656.18.408.1061525
968.335 76.48.16 0
168.615 76.5 8.28 144279.2 1 12 144
968.33576.48.160208.9656.18.408.1061525
Divide Equation 1 by 25 andmultiply it by 144, .
76.525
144
Subtract the result from Equation 3
Substitute new equation for Equation 3
![Page 24: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/24.jpg)
Forward Elimination: Step 2
728.33646.58.1605.3208.9656.18.40
968.33576.48.160208.9656.18.408.1061525
.760 7.0 0 0
728.33646.516.80335.968 76.416.80
76.07.000208.9656.18.408.1061525
Divide Equation 2 by −4.8and multiply it by −16.8, .
5.38.48.16
Subtract the result from Equation 3
Substitute new equation for Equation 3
![Page 25: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/25.jpg)
Back Substitution
![Page 26: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/26.jpg)
Back Substitution
76.0208.968.106
7.00056.18.40
1525
7.07.0002.9656.18.408.1061525
3
2
1
aaa
08571.17.076.076.07.0
3
3
3
a
a
aSolving for a3
![Page 27: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/27.jpg)
Back Substitution (cont.)
Solving for a2
690519. 4.8
1.085711.5696.208
8.456.1208.96
208.9656.18.4
2
2
32
32
a
a
aa
aa
76.0208.968.106
7.00056.18.40
1525
3
2
1
aaa
![Page 28: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/28.jpg)
Back Substitution (cont.)
Solving for a1
290472.025
08571.16905.1958.1062558.106
8.106525
321
321
aaa
aaa
76.02.968.106
7.00056.18.40
1525
3
2
1
aaa
![Page 29: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/29.jpg)
Naïve Gaussian Elimination Solution
227921778106
11214418641525
3
2
1
.
.
.
aaa
08571.16905.19
290472.0
3
2
1
aaa
![Page 30: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/30.jpg)
Example 1 Cont.SolutionThe solution vector is
08571.16905.19
290472.0
3
2
1
aaa
The polynomial that passes through the three data points is then:
125 ,08571.16905.19290472.0 2
322
1
ttt
atatatv
.m/s 686.129
08571.166905.196290472.06 2
v
![Page 31: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/31.jpg)
Naïve Gauss EliminationPitfalls
![Page 32: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/32.jpg)
95511326
3710
321
321
32
xxxxxx
xx
Pitfall#1. Division by zero
9113
5153267100
3
2
1
xxx
![Page 33: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/33.jpg)
Is division by zero an issue here?
95514356
1571012
321
321
321
xxxxxx
xxx
91415
51535671012
3
2
1
xxx
![Page 34: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/34.jpg)
Is division by zero an issue here? YES
2852414356
1571012
321
321
321
xxxxxx
xxx
281415
512435671012
3
2
1
xxx
25.6
15
1921125.60071012
3
2
1
xxx
Division by zero is a possibility at any step of forward elimination
![Page 35: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/35.jpg)
Pitfall#2. Large Round-off Errors
9751.145
3157249.23
101520
3
2
1
xxx
Exact Solution
111
3
2
1
xxx
![Page 36: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/36.jpg)
Pitfall#2. Large Round-off Errors
9751.145
3157249.23
101520
3
2
1
xxx
Solve it on a computer using 6 significant digits with chopping
999995.005.1
9625.0
3
2
1
xxx
![Page 37: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/37.jpg)
Pitfall#2. Large Round-off Errors
9751.145
3157249.23
101520
3
2
1
xxx
Solve it on a computer using 5 significant digits with chopping
99995.05.1
625.0
3
2
1
xxx
Is there a way to reduce the round off error?
![Page 38: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/38.jpg)
Avoiding PitfallsIncrease the number of significant digits• Decreases round-off error• Does not avoid division by zero
![Page 39: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/39.jpg)
Avoiding Pitfalls
Gaussian Elimination with Partial Pivoting
• Avoids division by zero• Reduces round off error
![Page 40: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/40.jpg)
THE END
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Gauss Elimination with Partial Pivoting
![Page 42: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/42.jpg)
Pitfalls of Naïve Gauss Elimination
• Possible division by zero• Large round-off errors
![Page 43: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/43.jpg)
Avoiding PitfallsIncrease the number of significant digits• Decreases round-off error• Does not avoid division by zero
![Page 44: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/44.jpg)
Avoiding PitfallsGaussian Elimination with Partial
Pivoting• Avoids division by zero• Reduces round off error
![Page 45: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/45.jpg)
What is Different About Partial Pivoting?
pka
At the beginning of the kth step of forward elimination, find the maximum of
nkkkkk aaa .......,,........., ,1
If the maximum of the values is in the p th row, ,npk then switch rows p
and k.
![Page 46: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/46.jpg)
Matrix Form at Beginning of 2nd Step of Forward Elimination
'
'3
2
1
3
2
1
''4
'32
'3
'3332
22322
1131211
0
00
n
'
nnnnn'n
n'
'n
''n
b
bbb
x
xxx
aaaa
aaaaaaaaaa
![Page 47: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/47.jpg)
Example (2nd step of FE)
3986
5
43111217086239011112402167067.31.5146
5
4
3
2
1
xxxxx
Which two rows would you switch?
![Page 48: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/48.jpg)
Example (2nd step of FE)
69835
21670862390111124043111217067.31.5146
5
4
3
2
1
xxxxx
Switched Rows
![Page 49: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/49.jpg)
Gaussian Elimination with Partial Pivoting
A method to solve simultaneous linear equations of the form [A][X]=[C]
Two steps1. Forward Elimination2. Back Substitution
![Page 50: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/50.jpg)
Forward EliminationSame as naïve Gauss elimination
method except that we switch rows before each of the (n-1) steps of forward elimination.
![Page 51: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/51.jpg)
Example: Matrix Form at Beginning of 2nd Step of
Forward Elimination
'
'3
2
1
3
2
1
''4
'32
'3
'3332
22322
1131211
0
00
n
'
nnnnn'n
n'
'n
''n
b
bbb
x
xxx
aaaa
aaaaaaaaaa
![Page 52: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/52.jpg)
Matrix Form at End of Forward Elimination
)(n-n
"
'
n)(n
nn
"n
"
'n
''n
b
bbb
x
xxx
a
aaaaaaaaa
1
3
2
1
3
2
1
1
333
22322
1131211
0000
000
![Page 53: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/53.jpg)
Back Substitution Starting Eqns
'2
'23
'232
'22 ... bxaxaxa nn
"3
"3
"33 ... bxaxa nn
11 nnn
nnn bxa
. .
. . . .
11313212111 ... bxaxaxaxa nn
![Page 54: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/54.jpg)
Back Substitution
1,...,1for11
11
nia
xabx i
ii
n
ijj
iij
ii
i
)1(
)1(
nnn
nn
n ab
x
![Page 55: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/55.jpg)
THE END
![Page 56: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/56.jpg)
Gauss Elimination with Partial Pivoting
Example
![Page 57: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/57.jpg)
Example 2
227921778106
11214418641525
3
2
1
.
.
.
aaa
Solve the following set of equations by Gaussian elimination with partial pivoting
![Page 58: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/58.jpg)
Example 2 Cont.
227921778106
11214418641525
3
2
1
.
.
.
aaa
1. Forward Elimination2. Back Substitution
2.2791121442.17718648.1061525
![Page 59: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/59.jpg)
Forward Elimination
![Page 60: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/60.jpg)
Number of Steps of Forward Elimination
Number of steps of forward elimination is (n1)=(31)=2
![Page 61: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/61.jpg)
Forward Elimination: Step 1• Examine absolute values of first column, first row
and below.144,64,25
• Largest absolute value is 144 and exists in row 3.
• Switch row 1 and row 3.
8.10615252.17718642.279112144
2.2791121442.17718648.1061525
![Page 62: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/62.jpg)
Forward Elimination: Step 1 (cont.)
.
1.1244444.0333.599.634444.02.279112144
8.10615252.17718642.279112144
10.53.55560667.2 0
124.1 0.44445.33363.99177.21 8 64
8.106152510.535556.0667.20
2.279112144
Divide Equation 1 by 144 andmultiply it by 64, .
4444.014464
Subtract the result from Equation 2
Substitute new equation for Equation 2
![Page 63: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/63.jpg)
Forward Elimination: Step 1 (cont.)
.
47.481736.0083.200.251736.0279.2112144
33.588264.0 917.20
48.470.17362.08325106.81 5 25
8.106152510.535556.0667.20
2.279112144
33.588264.0917.2010.535556.0667.20
2.279112144
Divide Equation 1 by 144 andmultiply it by 25, .
1736.014425
Subtract the result from Equation 3
Substitute new equation for Equation 3
![Page 64: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/64.jpg)
Forward Elimination: Step 2• Examine absolute values of second column, second
row and below. 2.917,667.2
• Largest absolute value is 2.917 and exists in row 3.
• Switch row 2 and row 3.
10.535556.0667.2033.588264.0917.20
2.279112144
33.588264.0917.2010.535556.0667.20
2.279112144
![Page 65: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/65.jpg)
Forward Elimination: Step 2 (cont.)
.
33.537556.0667.209143.058.330.82642.9170
10.535556.0667.2033.588264.0917.202.279112144
23.02.0 0 0
53.33 0.75562.667053.10 0.55562.6670
23.02.00033.588264.0917.202.279112144
Divide Equation 2 by 2.917 andmultiply it by 2.667, .9143.0
917.2667.2
Subtract the result from Equation 3
Substitute new equation for Equation 3
![Page 66: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/66.jpg)
Back Substitution
![Page 67: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/67.jpg)
Back Substitution
1.152.023.023.02.0
3
3
a
a
Solving for a3
2303358
2279
20008264091720112144
23.02.00033.588264.0917.20
2.279112144
3
2
1
...
aaa
.
..
![Page 68: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/68.jpg)
Back Substitution (cont.)
Solving for a2
6719.917.2
15.18264.033.58917.2
8264.033.5833.588264.0917.2
32
32
aa
aa
2303358
2279
20008264091720112144
3
2
1
...
aaa
.
..
![Page 69: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/69.jpg)
Back Substitution (cont.)
Solving for a1
2917.0144
15.167.19122.279144122.279
2.27912144
321
321
aaa
aaa
2303358
2279
20008264091720112144
3
2
1
...
aaa
.
..
![Page 70: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/70.jpg)
Gaussian Elimination with Partial Pivoting Solution
227921778106
11214418641525
3
2
1
.
.
.
aaa
15.167.19
2917.0
3
2
1
aaa
![Page 71: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/71.jpg)
Gauss Elimination with Partial Pivoting
Another Example
![Page 72: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/72.jpg)
Partial Pivoting: ExampleConsider the system of equations
655901.36099.23
7710
321
321
21
xxxxxx
xx
In matrix form
5156099.230710
3
2
1
xxx
6901.37
=
Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping
![Page 73: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/73.jpg)
Partial Pivoting: ExampleForward Elimination: Step 1
Examining the values of the first column
|10|, |-3|, and |5| or 10, 3, and 5
The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we switch row1 with row1.
6901.37
5156099.230710
3
2
1
xxx
5.2001.67
55.206001.000710
3
2
1
xxx
Performing Forward Elimination
![Page 74: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/74.jpg)
Partial Pivoting: ExampleForward Elimination: Step 2
Examining the values of the first column
|-0.001| and |2.5| or 0.0001 and 2.5
The largest absolute value is 2.5, so row 2 is switched with row 3
5.2001.67
55.206001.000710
3
2
1
xxx
001.65.2
7
6001.0055.200710
3
2
1
xxx
Performing the row swap
![Page 75: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/75.jpg)
Partial Pivoting: ExampleForward Elimination: Step 2
Performing the Forward Elimination results in:
002.65.2
7
002.60055.200710
3
2
1
xxx
![Page 76: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/76.jpg)
Partial Pivoting: ExampleBack Substitution
Solving the equations through back substitution
1002.6002.6
3 x
15.255.2 3
2
xx
010
077 321
xxx
002.65.2
7
002.60055.200710
3
2
1
xxx
![Page 77: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/77.jpg)
Partial Pivoting: Example
11
0
3
2
1
xxx
X exact
11
0
3
2
1
xxx
X calculated
Compare the calculated and exact solution
The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting
![Page 78: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/78.jpg)
THE END
![Page 79: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/79.jpg)
Determinant of a Square Matrix
Using Naïve Gauss EliminationExample
![Page 80: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/80.jpg)
Theorem of Determinants If a multiple of one row of [A]nxn is added or subtracted to another row of [A]nxn to result in [B]nxn then det(A)=det(B)
![Page 81: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/81.jpg)
Theorem of Determinants The determinant of an upper triangular matrix [A]nxn is given by
nnii aaaa ......Adet 2211
n
iiia
1
![Page 82: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/82.jpg)
Forward Elimination of a Square Matrix
Using forward elimination to transform [A]nxn
to an upper triangular matrix, [U]nxn. nnnn UA
UA detdet
![Page 83: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/83.jpg)
Example
11214418641525
Using naïve Gaussian elimination find the determinant of the following square matrix.
![Page 84: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/84.jpg)
Forward Elimination
![Page 85: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/85.jpg)
Forward Elimination: Step 1
.
56.28.126456.21525 56.18.40
56.2 8.12641 8 64
11214418641525
11214456.18.40
1525
Divide Equation 1 by 25 andmultiply it by 64, .
56.22564
Subtract the result from Equation 2
Substitute new equation for Equation 2
![Page 86: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/86.jpg)
Forward Elimination: Step 1 (cont.)
.
76.58.2814476.51525 76.48.16 0
76.5 8.281441 12 144
76.48.16056.18.40
1525
11214456.18.40
1525
Divide Equation 1 by 25 andmultiply it by 144, .
76.525
144
Subtract the result from Equation 3
Substitute new equation for Equation 3
![Page 87: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/87.jpg)
Forward Elimination: Step 2
.
76.48.16056.18.40
1525
46.58.1605.356.18.40
7.0 0 0
46.58.16076.48.160
7.00056.18.40
1525
Divide Equation 2 by −4.8and multiply it by −16.8, .
5.38.48.16
Subtract the result from Equation 3
Substitute new equation for Equation 3
![Page 88: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/88.jpg)
Finding the Determinant
.
7.00056.18.40
1525
11214418641525
After forward elimination
00.84
7.08.425Adet 332211
uuu
![Page 89: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/89.jpg)
Summary- Forward Elimination- Back Substitution- Pitfalls- Improvements- Partial Pivoting- Determinant of a Matrix
![Page 90: Gaussian Elimination](https://reader033.vdocument.in/reader033/viewer/2022050712/5681614b550346895dd0cd99/html5/thumbnails/90.jpg)
THE END