Major: All Engineering Majors Author(s): Autar KawForward Elimination = 279.2 177.2 106.8 144 12 1...
Transcript of Major: All Engineering Majors Author(s): Autar KawForward Elimination = 279.2 177.2 106.8 144 12 1...
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Gaussian Elimination
Major: All Engineering Majors
Author(s): Autar Kaw
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
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Naïve Gauss Elimination
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Naïve Gaussian EliminationA method to solve simultaneous linear equations of the form [A][X]=[C]
Two steps1. Forward Elimination2. Back Substitution
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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
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Forward EliminationA set of n equations and n unknowns
11313212111 ... bxaxaxaxa nn =++++
22323222121 ... bxaxaxaxa nn =++++
nnnnnnn bxaxaxaxa =++++ ...332211
. .
. .
. .
(n-1) steps of forward elimination
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Forward EliminationStep 1 For Equation 2, divide Equation 1 by and multiply by .
)...( 1131321211111
21 bxaxaxaxaaa
nn =++++
111
211
11
21212
11
21121 ... ba
axaaaxa
aaxa nn =+++
11a21a
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Forward Elimination
111
211
11
21212
11
21121 ... ba
axaaaxa
aaxa nn =+++
111
2121
11
212212
11
2122 ... ba
abxaaaaxa
aaa nnn −=
−++
−
'2
'22
'22 ... bxaxa nn =++
22323222121 ... bxaxaxaxa nn =++++Subtract the result from Equation 2.
−_________________________________________________
or
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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
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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
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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 −− = nnnn
nn bxa
. .. .. .
11313212111 ... bxaxaxaxa nn =++++
End of Step (n-1)
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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
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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
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Back Substitution Starting Eqns
'2
'23
'232
'22 ... bxaxaxa nn =+++
"3
"3
"33 ... bxaxa nn =++
( ) ( )11 −− = nnnn
nn bxa
. .. .. .
11313212111 ... bxaxaxaxa nn =++++
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Back SubstitutionStart with the last equation because it has only one unknown
)1(
)1(
−
−
= nnn
nn
n ab
x
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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
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THE END
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Additional Resources
For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit
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Naïve Gauss EliminationExample
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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 , 3221 ≤≤++= atatatvFind 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 ≤≤++= 3221
=
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
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Example 1 Cont.
⇒
=
22791121442177186481061525
227921778106
11214418641525
3
2
1
.
.
.
.
.
.
aaa
1. Forward Elimination2. Back Substitution
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Forward Elimination
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Number of Steps of Forward Elimination
Number of steps of forward elimination is (n−1)=(3−1)=2
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Divide Equation 1 by 25 and
multiply 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 and
multiply it by 144, .76.525
144=
Subtract the result from Equation 3
Substitute new equation for Equation 3
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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 −4.8
and multiply it by −16.8,
.5.38.48.16=
−−
Subtract the result from Equation 3
Substitute new equation for Equation 3
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Back Substitution
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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
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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
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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
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Naïve Gaussian Elimination Solution
=
227921778106
11214418641525
3
2
1
.
.
.
aaa
=
08571.16905.19
290472.0
3
2
1
aaa
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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
≤≤++=
++=
tttatatatv
( ) ( ) ( ).m/s 686.129
08571.166905.196290472.06 2
=++=v
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THE END
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Naïve Gauss EliminationPitfalls
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-
95511326
3710
321
321
32
=+−=++
=−
xxxxxx
xx
Pitfall#1. Division by zero
=
−
−
9113
5153267100
3
2
1
xxx
-
Is division by zero an issue here?
95514356
1571012
321
321
321
=+−=++=−+
xxxxxx
xxx
=
−
−
91415
51535671012
3
2
1
xxx
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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
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Pitfall#2. Large Round-off Errors
=
−−
9751.145
3157249.23
101520
3
2
1
xxx
Exact Solution
=
111
3
2
1
xxx
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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
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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?
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Avoiding Pitfalls
Increase the number of significant digits• Decreases round-off error
• Does not avoid division by zero
-
Avoiding Pitfalls
Gaussian Elimination with Partial Pivoting• Avoids division by zero
• Reduces round off error
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THE END
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Gauss Elimination with Partial Pivoting
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-
Pitfalls of Naïve Gauss Elimination
• Possible division by zero• Large round-off errors
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Avoiding Pitfalls
Increase the number of significant digits• Decreases round-off error
• Does not avoid division by zero
-
Avoiding Pitfalls
Gaussian Elimination with Partial Pivoting• Avoids division by zero
• Reduces round off error
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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.
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Matrix Form at Beginning of 2ndStep 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
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Example (2nd step of FE)
−
=
−
−
3986
5
43111217086239011112402167067.31.5146
5
4
3
2
1
xxxxx
Which two rows would you switch?
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Example (2nd step of FE)
−
=
−
−
69835
21670862390
111124043111217067.31.5146
5
4
3
2
1
xxxxx
Switched Rows
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Gaussian Elimination with Partial Pivoting
A method to solve simultaneous linear equations of the form [A][X]=[C]
Two steps1. Forward Elimination2. Back Substitution
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Forward Elimination
Same as naïve Gauss elimination method except that we switch rows before eachof the (n-1) steps of forward elimination.
-
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
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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
-
Back Substitution Starting Eqns
'2
'23
'232
'22 ... bxaxaxa nn =+++
"3
"3
"33 ... bxaxa nn =++
( ) ( )11 −− = nnnn
nn bxa
. .. .. .
11313212111 ... bxaxaxaxa nn =++++
-
Back Substitution
( ) ( )
( ) 1,...,1for11
11
−=∑−
= −+=
−−
nia
xabx i
ii
n
ijj
iij
ii
i
)1(
)1(
−
−
= nnn
nn
n ab
x
-
THE END
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Gauss Elimination with Partial Pivoting
Example
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-
Example 2
=
227921778106
11214418641525
3
2
1
.
.
.
aaa
Solve the following set of equations by Gaussian elimination with partial pivoting
-
Example 2 Cont.
⇒
=
227921778106
11214418641525
3
2
1
.
.
.
aaa
1. Forward Elimination2. Back Substitution
2.2791121442.17718648.1061525
-
Forward Elimination
-
Number of Steps of Forward Elimination
Number of steps of forward elimination is (n−1)=(3−1)=2
-
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
-
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 and
multiply it by 64, .4444.014464
=
Subtract the result from Equation 2
Substitute new equation for Equation 2
-
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 and
multiply it by 25, .1736.014425
=
Subtract the result from Equation 3
Substitute new equation for Equation 3
-
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
-
Forward Elimination: Step 2 (cont.)
.
[ ] [ ]33.537556.0667.209143.058.330.82642.9170 =×
10.535556.0667.2033.588264.0917.20
2.279112144
[ ][ ][ ]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,
.9143.0917.2667.2
=
Subtract the result from Equation 3
Substitute new equation for Equation 3
-
Back Substitution
-
Back Substitution
1.152.023.023.02.0
3
3
=−−
=
−=−
a
aSolving for a3
−=
−⇒
−− 2303358
2279
20008264091720112144
23.02.00033.588264.0917.202.279112144
3
2
1
...
aaa
.
..
-
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
.
..
-
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
.
..
-
Gaussian Elimination with Partial Pivoting Solution
=
227921778106
11214418641525
3
2
1
.
.
.
aaa
=
15.167.19
2917.0
3
2
1
aaa
-
Gauss Elimination with Partial Pivoting
Another Example
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-
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
-
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
-
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
-
Partial Pivoting: Example
Forward Elimination: Step 2
Performing the Forward Elimination results in:
=
−
002.65.2
7
002.60055.200710
3
2
1
xxx
-
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
-
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
-
THE END
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Determinant of a Square MatrixUsing Naïve Gauss Elimination
Example
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-
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)
-
Theorem of Determinants
The determinant of an upper triangular matrix
[A]nxn is given by
( ) nnii aaaa ×××××= ......Adet 2211
∏=
=n
iiia
1
-
Forward Elimination of a Square Matrix
Using forward elimination to transform [A]nxn to an
upper triangular matrix, [U]nxn.
[ ] [ ] nnnn UA ×× →
( ) ( )UA detdet =
-
Example
11214418641525
Using naïve Gaussian elimination find the determinant of the following square matrix.
-
Forward Elimination
-
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 and
multiply it by 64, .56.22564
=
Subtract the result from Equation 2
Substitute new equation for Equation 2
-
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 and
multiply it by 144, .76.525
144=
Subtract the result from Equation 3
Substitute new equation for Equation 3
-
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 −4.8
and multiply it by −16.8,
.5.38.48.16=
−−
Subtract the result from Equation 3
Substitute new equation for Equation 3
-
Finding the Determinant
.
7.00056.18.40
1525
11214418641525
−−→
After forward elimination
( )( )00.84
7.08.425Adet 332211
−=×−×=
××= uuu
-
Summary
-Forward Elimination
-Back Substitution
-Pitfalls
-Improvements
-Partial Pivoting
-Determinant of a Matrix
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Additional Resources
For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit
http://numericalmethods.eng.usf.edu/topics/gaussian_elimination.html
http://numericalmethods.eng.usf.edu/topics/gaussian_elimination.html
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THE END
http://numericalmethods.eng.usf.edu
http://numericalmethods.eng.usf.edu/
Gaussian EliminationNaïve Gauss Elimination�� ��http://numericalmethods.eng.usf.edu��Naïve Gaussian EliminationForward EliminationForward EliminationForward EliminationForward EliminationForward EliminationForward EliminationForward EliminationMatrix Form at End of Forward EliminationBack SubstitutionBack Substitution Starting EqnsBack SubstitutionBack SubstitutionSlide Number 16Additional ResourcesNaïve Gauss Elimination�Example���� http://numericalmethods.eng.usf.eduExample 1Example 1 Cont. �Example 1 Cont.�Forward EliminationNumber of Steps of Forward EliminationForward Elimination: Step 1Forward Elimination: Step 1 (cont.)Forward Elimination: Step 2Back SubstitutionBack SubstitutionBack Substitution (cont.)Back Substitution (cont.)Naïve Gaussian Elimination SolutionExample 1 Cont.Slide Number 33Naïve Gauss Elimination�Pitfalls��� http://numericalmethods.eng.usf.eduSlide Number 35Is division by zero an issue here?Is division by zero an issue here? YESSlide Number 38Slide Number 39Slide Number 40Avoiding PitfallsAvoiding PitfallsSlide Number 43Gauss Elimination with Partial Pivoting��� http://numericalmethods.eng.usf.eduPitfalls of Naïve Gauss EliminationAvoiding PitfallsAvoiding PitfallsWhat is Different About Partial Pivoting?Matrix Form at Beginning of 2nd Step of Forward EliminationExample (2nd step of FE)Example (2nd step of FE)Gaussian Elimination �with Partial PivotingForward EliminationExample: Matrix Form at Beginning of 2nd Step of Forward EliminationMatrix Form at End of Forward EliminationBack Substitution Starting EqnsBack SubstitutionSlide Number 58Gauss Elimination with Partial Pivoting�Example��� http://numericalmethods.eng.usf.edu�Example 2��Example 2 Cont.�Forward EliminationNumber of Steps of Forward EliminationForward Elimination: Step 1Forward Elimination: Step 1 (cont.)Forward Elimination: Step 1 (cont.)Forward Elimination: Step 2Forward Elimination: Step 2 (cont.)Back SubstitutionBack SubstitutionBack Substitution (cont.)Back Substitution (cont.)Gaussian Elimination with Partial Pivoting SolutionGauss Elimination with Partial Pivoting�Another Example��� http://numericalmethods.eng.usf.eduPartial Pivoting: ExamplePartial Pivoting: ExamplePartial Pivoting: ExamplePartial Pivoting: ExamplePartial Pivoting: ExamplePartial Pivoting: ExampleSlide Number 81Determinant of a Square Matrix�Using Naïve Gauss Elimination�Example���� http://numericalmethods.eng.usf.eduTheorem of Determinants Theorem of Determinants Forward Elimination of a �Square Matrix�Example�Forward EliminationForward Elimination: Step 1Forward Elimination: Step 1 (cont.)Forward Elimination: Step 2Finding the DeterminantSummaryAdditional ResourcesSlide Number 94