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![Page 1: 1 Solution of linear system of equations Circuit analysis (Mesh and node equations) Numerical solution of differential equations (Finite Difference Method)](https://reader030.fdocuments.net/reader030/viewer/2022033018/56649d215503460f949f5f9d/html5/thumbnails/1.jpg)
1
Solution of linear system of equations
Circuit analysis (Mesh and node equations)
Numerical solution of differential equations (Finite Difference Method)
Numerical solution of integral equations (Finite Element Method, Method of Moments)
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111
nnnnnn
n
n
b
b
b
x
x
x
aaa
aaa
aaa
2
1
2
1
21
22221
11211
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2
Consistency (Solvability)
The linear system of equations Ax=b has a solution, or said to be consistent IFF
Rank{A}=Rank{A|b} A system is inconsistent when
Rank{A}<Rank{A|b}
Rank{A} is the maximum number of linearly independent columns or rows of A. Rank can be found by using ERO (Elementary Row Oparations) or ECO (Elementary column operations).
ERO# of rows with at least one nonzero entryECO# of columns with at least one nonzero entry
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3
Elementary row operations
The following operations applied to the augmented matrix [A|b], yield an equivalent linear system Interchanges: The order of two rows can be
changed
Scaling: Multiplying a row by a nonzero constant
Replacement: The row can be replaced by the sum of that row and a nonzero multiple of any other row.
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4
An inconsistent example
5
4
42
21
2
1
x
x
00
21Rank{A}=1
Rank{A|b}=2
ERO:Multiply the first row with -2 and add to the second row
34
0
2
0
1
Then this system of equations
is not solvable
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5
Uniqueness of solutions
The system has a unique solution IFF
Rank{A}=Rank{A|b}=n
n is the order of the system
Such systems are called full-rank systems
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6
Full-rank systems If Rank{A}=n
Det{A} 0 A is nonsingular so invertibleUnique solution
2
4
11
21
2
1
x
x
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7
Rank deficient matrices
If Rank{A}=m<nDet{A} = 0 A is singular so not invertible infinite number of solutions (n-m free variables)
under-determined system
8
4
42
21
2
1
x
x
Consistent so solvable
Rank{A}=Rank{A|b}=1
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8
Ill-conditioned system of equations
A small deviation in the entries of A matrix, causes a large deviation in the solution.
47.1
3
99.048.0
21
2
1
x
x
47.1
3
99.049.0
21
2
1
x
x
1
1
2
1
x
x
0
3
2
1
x
x
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9
Ill-conditioned continued.....
A linear system
of equations is
said to be “ill-
conditioned” if
the coefficient
matrix tends to
be singular
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10
Types of linear system of equations to be studied in this course
Coefficient matrix A is square and real
The RHS vector b is nonzero and real
Consistent system, solvable
Full-rank system, unique solution
Well-conditioned system
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11
Solution Techniques Direct solution methods
Finds a solution in a finite number of operations by transforming the system into an equivalent system that is ‘easier’ to solve.
Diagonal, upper or lower triangular systems are easier to solve
Number of operations is a function of system size n.
Iterative solution methods Computes succesive approximations of the
solution vector for a given A and b, starting from an initial point x0.
Total number of operations is uncertain, may not converge.
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12
Direct solution Methods
Gaussian Elimination By using ERO, matrix A is transformed into an
upper triangular matrix (all elements below diagonal 0)
Back substitution is used to solve the upper-triangular system
n
i
n
i
nnnin
iniii
ni
b
b
b
x
x
x
aaa
aaa
aaa
11
1
1
1111
ERO
n
i
n
i
nn
inii
ni
b
b
b
x
x
x
a
aa
aaa
~
~
~00
~~0
111111
Back
sub
stit
uti
on
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13
First step of elimination
)2(
)2(3
)2(2
)1(1
3
2
1
)2()2(3
)2(2
)2(3
)2(33
)2(32
)2(2
)2(23
)2(22
)1(1
)1(13
)1(12
)1(11
)1(11
)1(11,
)1(11
)1(311,3
)1(11
)1(211,2
0
0
0
/
/
/
nnnnnn
n
n
n
nn b
b
b
b
x
x
x
x
aaa
aaa
aaa
aaaa
aam
aam
aam
)1(
)1(3
)1(2
)1(1
3
2
1
)1()1(3
)1(2
)1(1
)1(3
)1(33
)1(32
)1(31
)1(2
)1(23
)1(22
)1(21
)1(1
)1(13
)1(12
)1(11
nnnnnnn
n
n
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaa
aaaa
Pivotal element
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14
Second step of elimination
)3(
)3(3
)2(2
)1(1
3
2
1
)3()3(3
)3(3
)3(33
)2(2
)2(23
)2(22
)1(1
)1(13
)1(12
)1(11
)2(22
)2(22,
)2(22
)2(322,3
00
00
0
/
/
nnnnn
n
n
n
nn b
b
b
b
x
x
x
x
aa
aa
aaa
aaaa
aam
aam
)2(
)2(3
)2(2
)1(1
3
2
1
)2()2(3
)2(2
)2(3
)2(33
)2(32
)2(2
)2(23
)2(22
)1(1
)1(13
)1(12
)1(11
0
0
0
nnnnnn
n
n
n
b
b
b
b
x
x
x
x
aaa
aaa
aaa
aaaa
Pivotal element
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15
Gaussion elimination algorithm
Define number of steps as p (pivotal row) For p=1,n-1
For r=p+1 to n
For c=p+1 to n
0
/)(
)()(,
prp
ppp
prppr
a
aam
)(,
)()1( ppcpr
prc
prc amaa
)(,
)()1( pppr
pr
pr bmbb
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16
Back substitution algorithm
)(
)1(1
)3(3
)2(2
)1(1
1
3
2
1
)(
)(1
)(11
)3(3
)3(33
)2(2
)2(23
)2(22
)1(1
)1(13
)1(12
)1(11
0000
000
00
0
nn
nn
n
n
nnn
nnn
nnn
n
n
n
b
b
b
b
b
x
x
x
x
x
a
aa
aa
aaa
aaaa
1,,2,11
1
1
)()()(
11
)1(1)1(
111)(
)(
nnixaba
x
xaba
xa
bx
n
ikk
iik
iii
iii
nnnn
nnn
nnnn
nn
nn
n
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17
Operation count Number of arithmetic operations required
by the algorithm to complete its task. Generally only multiplications and
divisions are counted Elimination process
Back substitution
Total
6
5
23
23 nnn
2
2 nn
332
3 nn
n
DominatesNot efficient for
different RHS vectors
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18
LU Decomposition
A=LU
Ax=b LUx=b
Define Ux=y
Ly=b Solve y by forward substitution
ERO’s must be performed on b as well as A
The information about the ERO’s are stored in L
Indeed y is obtained by applying ERO’s to b vector
Ux=y Solve x by backward substitution
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19
LU Decomposition by Gaussian elimination
)(
)(1
)(11
)3(3
)3(33
)2(2
)2(23
)2(22
)1(1
)1(13
)1(12
)1(11
4,3,2,1,
3,12,11,1
2,31,3
1,2
0000
000
00
0
1
1
0
001
0001
00001
nnn
nnn
nnn
n
n
n
nnnn
nnn
a
aa
aa
aaa
aaaa
mmmm
mmm
mm
m
A
Compact storage: The diagonal entries of L matrix are all 1’s, they don’t need to be stored. LU is stored in a single matrix.
There are infinitely many different ways to decompose A.Most popular one: U=Gaussian eliminated matrix
L=Multipliers used for elimination
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20
Operation count
A=LU Decomposition
Ly=b forward substitution
Ux=y backward substitution
Total For different RHS vectors, the system can
be efficiently solved.
33
3 nn
2
2 nn 2
2 nn
332
3 nn
n
Done only once
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21
Pivoting Computer uses finite-precision arithmetic A small error is introduced in each arithmetic
operation, error propagates When the pivotal element is very small, the
multipliers will be large. Adding numbers of widely differening
magnitude can lead to loss of significance. To reduce error, row interchanges are made
to maximise the magnitude of the pivotal element
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22
Example: Without Pivoting
93.22
414.6
210.114.24
281.5133.1
2
1
x
x
8.113
414.6
7.113000.0
281.5133.1
2
1
x
x
001.1
9956.0
2
1
x
x
31.21133.1
14.2421 m
4-digit arithmetic
Loss of significance
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23
Example: With Pivoting
414.6
93.22
281.5133.1
210.114.24
2
1
x
x
338.5
93.22
338.5000.0
210.114.24
2
1
x
x
000.1
000.1
2
1
x
x
04693.014.24
133.121 m
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24
Pivoting procedures
)()()(
)()()(
)()()(
)3(3
)3(3
)3(3
)3(33
)2(2
)2(2
)2(2
)2(23
)2(22
)1(1
)1(1
)1(1
)1(13
)1(12
)1(11
000
000
000
00
0
inn
inj
ini
ijn
ijj
iji
iin
iij
iii
nji
nji
nji
aaa
aaa
aaa
aaaa
aaaaa
aaaaaa
Eliminated part
Pivotal column
Pivotal row
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25
Row pivoting
Most commonly used partial pivoting
procedure
Search the pivotal column
Find the largest element in magnitude
Then switch this row with the pivotal row
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26
Row pivoting
)()()(
)()()(
)()()(
)3(3
)3(3
)3(3
)3(33
)2(2
)2(2
)2(2
)2(23
)2(22
)1(1
)1(1
)1(1
)1(13
)1(12
)1(11
000
000
000
00
0
inn
inj
ini
ijn
ijj
iji
iin
iij
iii
nji
nji
nji
aaa
aaa
aaa
aaaa
aaaaa
aaaaaa
Interchange these rows
Largest in magnitude
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27
Column pivoting
)()()(
)()()(
)()()(
)3(3
)3(3
)3(3
)3(33
)2(2
)2(2
)2(2
)2(23
)2(22
)1(1
)1(1
)1(1
)1(13
)1(12
)1(11
000
000
000
00
0
inn
inj
ini
ijn
ijj
iji
iin
iij
iii
nji
nji
nji
aaa
aaa
aaa
aaaa
aaaaa
aaaaaa
Interchange these columns
Largest in magnitude
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28
Complete pivoting
)()()(
)()()(
)()()(
)3(3
)3(3
)3(3
)3(33
)2(2
)2(2
)2(2
)2(23
)2(22
)1(1
)1(1
)1(1
)1(13
)1(12
)1(11
000
000
000
00
0
inn
inj
ini
ijn
ijj
iji
iin
iij
iii
nji
nji
nji
aaa
aaa
aaa
aaaa
aaaaa
aaaaaa
Largest in magnitude
Interchange these columns
Interchange these rows
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29
Row Pivoting in LU Decomposition
When two rows of A are interchanged, those rows of b should also be interchanged.
Use a pivot vector. Initial pivot vector is integers from 1 to n.
When two rows (i and j) of A are interchanged, apply that to pivot vector.
n
i
jp
3
2
1
n
j
ip
3
2
1
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30
Modifying the b vector
When LU decomposition of A is done, the pivot vector tells the order of rows after interchanges
Before applying forward substitution to solve Ly=b, modify the order of b vector according to the entries of pivot vector
9
5
7
6
8
4
2
3
1
p
9.6
5.3
7.2
2.5
6.9
8.4
2.1
6.8
3.7
b
9.6
6.9
7.2
2.5
5.3
8.4
6.8
2.1
3.7
b
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31
LU decomposition algorithm with row pivoting
For k=1,n-1 column to be eliminatedp=k For r=k+1 to n if if p>k then For c=1 to n For r=k+1 to n
For c=k+1 to nkr
krk
kkk
krkkr
ma
aam
,)1(
)()(, /
)(,
)()1( kkckr
krc
krc amaa
rp then pkrk aa
taaaat pcpckckc ,,
Column search for maximum entry
Interchaning the rows
Updating L matrix
Updating U matrix
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32
Example
3
2
1
3
5
12
241
124
230
pbA
3
1
2
241
230
124
pA
Column search: Maximum magnitude second rowInterchange 1st and 2nd rows
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33
Example continued...
3
1
2
241
230
124
pA
Eliminate a21 and a31 by using a11 as pivotal
elementA=LU in compact form (in a single matrix)
3
1
2
75.15.325.0
230
124
pA
Multipliers (L matrix)
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34
Example continued...
1
3
2
230
75.15.325.0
124
pA
3
1
2
75.15.325.0
230
124
pA
Column search: Maximum magnitude at the third rowInterchange 2nd and 3rd rows
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35
Example continued...
1
3
2
230
75.15.325.0
124
pA
Eliminate a32 by using a22 as pivotal element
1
3
2
5.35.3/30
75.15.325.0
124
pA
Multipliers (L matrix)
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36
Example continued...
1
3
2
5.300
75.15.30
124
15.3/30
0125.0
001
pA
12
3
5
3
5
12
1
3
2
bbp
A’x=b’ LUx=b’ Ux=y
Ly=b’
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Example continued...
3
2
1
3
2
1
x
x
x
Backwardsubstitution
5.10
75.1
5
3
2
1
y
y
y
Forwardsubstitution
5.10
75.1
5
5.300
75.15.30
124
3
2
1
x
x
x
Ux=y
12
3
5
15.3/30
0125.0
001
3
2
1
y
y
y
Ly=b’
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Gauss-Jordan elimination The elements above the diagonal are made zero
at the same time that zeros are created below the diagonal
)1()1()1(2
)1(1
)1(2
)1(2
)1(22
)1(21
)1(1
)1(1
)1(12
)1(11
nnnnn
n
n
baaa
baaa
baaa
)2()2()2(2
)2(2
)2(2
)2(22
)1(1
)1(1
)1(12
)1(11
0
0
nnnn
n
n
baa
baa
baaa
)()(
)1(2
)2(22
)1(1
)1(11
00
00
00
nn
nnn
n
n
ba
ba
ba
)3()3(
)2(2
)2()2(22
)2(1
)2()1(11
00
0
0
nnn
nn
nn
ba
baa
baa
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39
Gauss-Jordan Elimination
Almost 50% more arithmetic operations than Gaussian elimination
Gauss-Jordan (GJ) Elimination is prefered when the inverse of a matrix is required.
Apply GJ elimination to convert A into an identity matrix.
IA
1AI
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Different forms of LU factorization
Doolittle formObtained by
Gaussian elimination
Crout form
Cholesky form
33
2322
131211
3231
21
333231
232221
131211
00
0
1
01
001
u
uu
uuu
ll
l
aaa
aaa
aaa
100
10
1
0
00
23
1312
333231
2221
11
333231
232221
131211
u
uu
lll
ll
l
aaa
aaa
aaa
100
10
1
00
00
00
1
01
001
23
1312
33
22
11
3231
21 u
uu
d
d
d
ll
l
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Crout form
First column of L is computed
Then first row of U is computed
The columns of L and rows of U are computed alternately
11 ii al
11
11 l
au jj
njjil
ulau
niijulal
ii
i
k kjikijij
j
kkjikijij
,,3,2,
,,2,1,
1
1
1
1
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Crout reduction sequence
1000
100
10
1
0
00
000
34
2423
141312
44434241
333231
2221
11
44434241
34333231
24232221
14131211
u
uu
uuu
llll
lll
ll
l
aaaa
aaaa
aaaa
aaaa
An entry of A matrix is used only once to compute the corresponding entry of L or U matrixSo columns of L and rows of U can be stored in A matrix
1
2
3
4
5
6
7
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Cholesky form
A=LDU (Diagonals of L and U are 1)
If A is symmetric
L=UT A= UT DU= UT D1/2D1/2U
U’= D1/2U A= U’T U’
This factorization is also called square root factorization. Only U’ need to be stored
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44
Solution of Complex Linear System of Equations
Cz=w
C=A+jB Z=x+jy w=u+jv
(A+jB)(x+jy)=(u+jv)
(Ax-By)+j(Bx+Ay)=u+jv
v
u
y
x
AB
BA
Real linear system of equations
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Large and Sparse Systems
When the linear system is large and sparse (a lot of zero entries), direct methods become inefficient due to fill-in terms.
Fill-in terms are those which turn out to be nonzero during elimination
5553
444241
353331
2422
141311
000
00
00
000
00
aa
aaa
aaa
aa
aaa
55
4544
353433
2422
141311
0000
000
00
000
00
a
aa
aaa
aa
aaa
Elimination
Fill-interms
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Sparse Matrices Node equation matrix is a sparse matrix. Sparse matrices are stored very efficiently
by storing only the nonzero entries When the system is very large (n=10,000)
the fill-in terms considerable increases storage requirements
In such cases iterative solution methods should be prefered instead of direct solution methods