engg 407_p12_l20_lecture_03
DESCRIPTION
Engg 407 Spring lecture 3TRANSCRIPT
ENGG 407
Numerical Methods in Engineering
P12L20
Lecture #3
Dr. Sameh Nassar
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Recall: ENGG 407 W12 Topics
1. Introduction and Mathematical Background (Ch. #1, #2)
2. “Roots of ” Nonlinear Equations (Ch. #3)
3. Linear Equations and Systems (Ch. #4)
© Sameh Nassar
4. Interpolation, Least-squares Estimation and Curve Fitting (Ch. #5)
5. Numerical Differentiation (Ch. #6)
6. Numerical Integration (Ch. #7)
7. Ordinary Differential Equations: Initial Value Problems (Ch. #8)
8. Ordinary Differential Equations: Boundary Value Problems (Ch. #9)
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Linear Algebraic Equations
• Simultaneous linear algebraic equations:
1n1n313212111
bxa....xaxaxa
bxa....xaxaxa
=++++
=++++ a’s:
Constant
Coefficients
© Sameh Nassar
nnnn3n32n21n1
2n2n323222121
bxa....xaxaxa
bxa....xaxaxa
=++++
=++++
MM
MM
Coefficients
b’s:
Constants
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
• In Matrix form:
{B}[A]{x} = or simply bAx =
Linear Algebraic Equations
© Sameh Nassar
[A] n*n square matrix of coefficients
{B} n*1 column vector of constants
{x} n*1 column vector of unknowns
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
2n232221
1n131211
a......aaa
a......aaa
2
1
x
x
2
1
b
b
Linear Algebraic Equations
© Sameh Nassar
=
nnn3n2n1
3n333231
2n232221
a......aaa
a......aaa
MMMMM
A
=
n
3
2
x
x
M
x
=
n
3
2
b
b
M
b
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Small Number of Equations
Methods:
• Small sets of simultaneous equations means in general that: n ≤ 3.
© Sameh Nassar
Methods:
(1) Graphical method
(2) Cramer’s rule
(3) Elimination of unknowns
Something you know or
may tried before
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Large Number of Equations
Methods:
(1) Gauss Elimination
• Means that we have to use “Numerical Methods”.
© Sameh Nassar
(1) Gauss Elimination
(2) Gauss-Jordan Elimination
(3) LU Decomposition
(4) Jacobi
(4) Gauss-Seidel
Direct (Elimination) Methods
Iterative methods
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Graphical Method (1/2)
Case of 2 Equations (for example), i.e. n =2:
© Sameh Nassar
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Graphical Methods (2/2)
)2(2x2x
)1(18x2x3
21
21
=+−
=+
© Sameh Nassar
)a2(1x2
1x
)a1(9x2
3x
12
12
+=
+−=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Cramer’s Rule
Case of 3 Equations (for example), i.e. n =3:
© Sameh Nassar
||
aab
aab
aab
x 33323
23222
13121
1 A=
||
baa
baa
baa
x 33231
22221
11211
3 A=
||
aba
aba
aba
x 33331
23221
13111
2 A=
Where | | is the corresponding Determinant
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Elimination of Unknowns
Case of 2 Equations (for example), i.e. n =2:
)2(2x2x
)1(18x2x3
21
21
=+−
=+
© Sameh Nassar
)b2(16x4218x3x 111 −=−⇒=+−− 4x1 =
)a2(2)9x2
3(2x
)a1(9x2
3x
11
12
=+−+−
+−=
394*2
39x
2
3x 12 =+−=+−=
x2 elimination
x1 back-substitution
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Problems with Some Linear Equations
© Sameh Nassar
No SolutionInfinite No. of Solutions
Singular Systems
Ill-Conditioned Systems
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
• It is an extension of the method ofElimination of Unknowns to large sets of equations by developing a systematic scheme or algorithm to eliminate
Naïve Gauss Elimination (or simply Gauss Elimination)
© Sameh Nassar
scheme or algorithm to eliminate unknowns and then back substitute.
• Therefore, as in the shown case of the solution of 2 equations, the technique for n equations consists of two stages (phases):
- Forward elimination of unknowns
- Back substitution
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
Case of 3 Equations (for example), i.e. n =3:
= 2
1
b
b
b
© Sameh Nassar
=
3
2
b
bb
3333231
2232221
1131211
baaa
baaa
baaa
M
M
M
Augmented Matrix
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
3333231
2232221
1131211
baaa
baaa
baaa
M
M
M
111
212 R)
a
a(R −
131
3 R)a
a(R −
Phase 1:Forward Elimination
© Sameh Nassar
3333231
333
22322
1131211
ba00
baa0
baaa
""
'''
M
M
M
33332
22322
1131211
baa0
baa0
baaa
'''
'''
M
M
M
111
3 a
222
323 R)
a
a(R
''−
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
33
3
a
b
""=3x
Phase 2: Back Substitution
1131211 baaa M
© Sameh Nassar
33a"
11
12131
a
aab 231
xxx
−−=
22
232
a
ab
''' 3
2
xx
−=
333
22322
ba00
baa0
""
'''
M
M
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
Pivot Equation
Pivot Coefficientor
© Sameh Nassar
or Pivot Element
Division of a row Ri by its pivot element (aii , a’ii ,
etc.), is called: Row Normalization
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
"""
''''
n
n
11n131211
baa
baaa
baaaa
M
M
M
......00
......0
......
3333
222322
(1) Forward Elimination
In General:
© Sameh Nassar
−− 1)(n1)(n
nnn ba M
MMMMMM
......000
......
1,....,2,11 −−=−
= −+=
−− ∑nnifor
a
xab
xii
n
ijjiji
i 1)(i
1)(i1)(i
1)-(n
1)-(n
nn
nn a
bx =
(2) Back Substitution “Upper” Triangular Matrix
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #1
Use Gauss Elimination to solve:
© Sameh Nassar
Check your answers by substituting them into the
original equations.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #1 (Sol. “1/3”)
−−5.62*)
10(4 =−−
© Sameh Nassar
5.02*)8
10(2 −=−
0.52*)8
12(2 =−−
−−
62212
44210
2228
M
M
M
Augmented Matrix
12 R)8
10(R −
13 R)8
12(R −
5.62*)8
10(4 =−−
5.62*)8
(4 =−−
0.12*)8
12(2 −=−
0.92*)8
12(6 =−−
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #1 (Sol. “2/3”)
−−−
5.65.65.00
2228
M
M
© Sameh Nassar
85.6*)5.0
0.1(5 −=−−−
45.6*)5.0
0.1(9 −=−−−
− 9510 M23 R)5.0
0.1(R−−−
−−−
−−
0.40.800
5.65.65.00
2228
M
M
M
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #1 (Sol. “3/3”)
5.08
4x3 =
−−=
−−−
−−
0.40.800
5.65.65.00
2228
M
M
M
5.65.0
5.0*5.65.6x2 −=
−−=
© Sameh Nassar
5.18
5.6*2)5.0*2(2x1 =−−−−−=
6)5.0(2)5.6(2)5.1(12
4)5.0(4)5.6(2)5.1(10
2)5.0(2)5.6(2)5.1(8
=+−+
=+−+
−=−−+
Check:
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
(1) Round-off Errors:become important when large numbers of equations are solved. Everyresult is dependent on previous results. Hence, an error in the earlysteps will tend to propagate.
Problems Encountered
© Sameh Nassar
(2) Ill-Conditioned Systems:are those where small changes in coefficients results in large changesin the solution.
(3) Division by Zero:For example if a11 is zero, the elimination procedure fails. On the otherhand, although there may be no zeros on the main diagonal in theoriginal matrix, the elimination process may create zero aii elements on
the main diagonal in the final matrix.
Similar problems may occur if a11 (or any resulted aii) is close to zero.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
(1) Use of More Significant Figures:To help in solving the problems of:• Round-off errors• Ill-conditioned systems
Overcoming Encountered Problems (1/2)
© Sameh Nassar
• Ill-conditioned systems
(2) Pivoting:Before each row is normalized, the largest available coefficient in thecolumn below the pivot element is determined. The rows can then beswitched so that the largest element is the pivot element. Thisapproach is called “Partial Pivoting”.
Ex:
−−
62212
44210
2228
M
M
M
−− 2228
44210
62212
M
M
M
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss Elimination Method
(2) “Partial” Pivoting:• Avoids division by zero• Minimizes Round-off errors• Serves as a partial remedy
Overcoming Encountered Problems (2/2)
© Sameh Nassar
• Serves as a partial remedy for Ill-conditioning.
(3) Scaling:When the magnitude of the elements of one or more of the equationsare greatly different from the magnitudes of the elements of the otherequations, this leads to significant round-off errors .
Scaling is accomplished by dividing the elements of each row by thelargest element in the row (thus, the resulted largest element is 1.0).
After scaling, pivoting is employed. (This is known as scaled pivoting).
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Scaling and Partial Pivoting (1/2)
−
−
01.0152.03.0
67.09.115.0
44.05.03.01.0
M
M
M
−−
01.0152.03.0
44.05.03.01.0
67.09.115.0
M
M
MPivoting only,
no scaling
9.1byRdivide
5.0byRdivide 1
Scaling
−
0.352610.52630.2632
88.016.02.0
M
M
−
67.09.115.0
44.05.03.01.0
M
M
© Sameh Nassar
−
0.01-152.03.0
0.352610.52630.2632
88.016.02.0
M
M
M
− 88.016.02.0
0.352610.52630.2632
0.01-152.03.0
M
M
MScaling, thenpivoting using
scaled coefficients
Scaling, then pivotingusing original coefficients(Recommended approach)
−
−
01.0152.03.0
67.09.115.0
44.05.03.01.0
M
M
M
−
−
44.05.03.01.0
67.09.115.0
01.0152.03.0
M
M
M
1byRdivide
9.1byRdivide
3
2Scaling
0.01-152.03.0
0.352610.52630.2632
M
M
− 01.0152.03.0
67.09.115.0
M
M
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Scaling and Partial Pivoting (2/2)
• As shown, scaling itself will sometimes introduce round-off errors.
• Thus, in most Gauss elimination applications, scaling is employed
© Sameh Nassar
applications, scaling is employed to calculate scaled values of the coefficients solely as a criterion for pivoting but the original coefficient values are retained for the actual elimination and substitution computations; details on previous slide!
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #2
Given the equations:
347xx3x
38x6x2x
321
321
−=+−−
−=−−
© Sameh Nassar
(a) Solve by Gauss elimination with partial pivoting only (i.e. no scaling). Show all steps of the computations.
(b) Substitute your results into the original equations to check your answers.
202xx8x 321 −=−+−
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #2 (Sol. “1/3”)
202xx8x
347xx3x
38x6x2x
321
321
321
−=−+−
−=+−−
−=−−
−−−−−−−−−
20218
34713
38162
M
M
MAugmented
Matrix
5.2620*)3
(34 −=−−−−
© Sameh Nassar
−−−−−−−−−
38162
34713
20218
M
M
M
-1.3751*)8
3(1 =−−−−
5.12*)8
2(1 −=−−
−−
12 )8
3( RR−−−
13 )8
2( RR−
−
75.72*)8
3(7 =−−−−
-5.751*)8
2(6 =−
−−
5.2620*)8
(34 −=−−
−−
4320*)8
2(38 −=−−
−−
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #2 (Sol. “2/3”)
−−−−−−−−
435.175.50
5.2675.7375.10
20218
M
M
M
-16.217443*)75.5
375.1(5.26 =−
−−−−
© Sameh Nassar
−−−−−−−−
5.2675.7375.10
435.175.50
20218
M
M
M
8.1086965.1*)75.5
375.1(75.7 =−
−−−
23 )75.5
375.1( RR
−−−
−−−−−−
16.2174-8.10869600
435.175.50
20218
M
M
M
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #2 (Sol. “3/3”)
28.108696
16.2174-x3 −==
875.5
2*)5.1(432 =
−−−−−=x
−−−−−−
16.2174-8.10869600
435.175.50
20218
M
M
M
© Sameh Nassar
75.5−
48
8*12*)2(20x1 =
−−−−−−=
Check:
20(-2)*284*8
34(*784*3
38(8*64*2
−=−+−
−=−+−−
−=−−−
)2
)2
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss-Jordan Elimination
• Is a variation of Gauss Elimination.
• The major difference is that when an unknown is eliminated in the Gauss-Jordan Elimination method, it is eliminated from all other equations rather than just the subsequent ones as in
© Sameh Nassar
all other equations rather than just the subsequent ones as in Gauss elimination.
• In addition, all rows are normalized by dividing them by their pivot elements.
• Thus, the elimination step results in an Identity matrixrather than an upper triangular matrix.
• Hence, it is not necessary to employ back substitution.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss-Jordan Elimination
© Sameh Nassar
Case of n = 3
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #3
Use Gauss-Jordan elimination to solve:
5
1
−=++
=−+ 321
42x2xx
xx2x
© Sameh Nassar
Do not employ pivoting. Check your answers by substituting
them into the original equations.
53
5
=++
−=++
321
321
xxx
42x2xx
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #2 (Sol. “1/4”)
−−
5113
4225
1112
M
M
M
53
5
1
=++
−=++
=−+
321
321
321
xxx
42x2xx
xx2x
−−
5113
4225
5.05.05.01
M
M
MR1/2
© Sameh Nassar
−−
5113
4225
5.05.05.01
M
M
M
-0.55.0*52 =−
5.25.0*31 =−−
12 *5 RR −
13 *3 RR −
5.45.0*52 =−−
.50-5.0*31 =−
5.65.0*54 −=−−
5.35.0*35 =−
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #3 (Sol. “2/4”)
−−−
−
5.35.25.00
5.65.45.00
5.05.05.01
M
M
M
−−
−
5.35.25.00
13910
5.05.05.01
M
M
M
R2/-0.5
49*)5.0(5.0 =−−−
© Sameh Nassar
−−
−
5.35.25.00
13910
5.05.05.01
M
M
M
-29*)5.0(5.2 =−−−
1013*)5.0(5.3 =−−
23 *)5.0( RR −−
−−
102-00
13910
6401
M
M
M
21 *5.0 RR −
613*5.05.0 −=−
49*)5.0(5.0 =−−−
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #3 (Sol. “3/4”)
−−
−
5100
13910
6401
M
M
M
R3/-2
−−
102-00
13910
6401
M
M
M
© Sameh Nassar
−−
−
5100
13910
6401
M
M
M
235*)9(13 −=−−−
32 )9( RR −−
−5-100
32010
14001
M
M
M
31 *4 RR −145*46 =−−−
X1 = 14
X2 = -32
X3 = -5
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #3 (Sol. “4/4”)
Check:
1=−+ (-5)(-32)14*2
© Sameh Nassar
53
5
=++
−=++
(-5)(-32)14*
4(-5)*2(-32)*214*
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
• LU decomposition is a class of elimination methods
• Its major appeal is that the time-consuming elimination step can be formulated so that it involves only operations on the matrix of coefficients A.
© Sameh Nassar
operations on the matrix of coefficients A.
• Therefore, it is will suited for situations where many right-hand-size vectors {B} (i.e. b) must be evaluated for a single value of A.
• LU decomposition provides an efficient means for computing the matrix inverse, i.e. A-1.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
Define:
Then:
© Sameh Nassar
Similar to the 1st phase of Gauss elimination, consider:
Then:
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
© Sameh Nassar
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
General Forms (for computing the elements of L and U):
Computing the 1st column of L (for known u11) and 1st row of U (for known l11)
© Sameh Nassar
Computing the elements of the ith row of U or the jth column of L
Computing the main diagonal elements of L (for known uii) and U (for known l ii)
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
General Solution (for d and x):
dlbi 1
−∑−
In General:
b
© Sameh Nassar
Forward Substitution
Back Substitution1,....,2,11 −−=
−=
∑+= nnifor
u
xud
xii
n
ijjiji
inn
nn u
dx =
niforl
dlb
dii
jjiji
i ,....,3,21 =−
=∑
=
11
11 l
bd =
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
• Thus, the LU decomposition method is decomposing (factoring) the coefficient matrix A into a product of 2 matrices L (lower-triangular matrix) and U (upper-triangular matrix).
© Sameh Nassar
• Several methods can be used to obtain the corresponding L and U.
Methods to be discussed:
(1) Doolittle Decomposition (all l ii “i.e. L main diagonal elements” = 1).
(2) Crout’s method all uii “i.e. U main diagonal elements” = 1).
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
• Is related to Gauss Elimination.
(1) Doolittle Decomposition (Factorization):
131211 aaa
Case of 3 Equations (forex.), i.e. n =3:
© Sameh Nassar
333231
232221
131211
aaa
aaa
aaa
33
2322
a"00
a'a'0
22
3232
11
3131
11
2121 a'
a'l
a
al
a
al ===
1ll01l001
3231
21
Using Gauss Elimination:
A
U
L
A = LU
Coefficients of Gauss elimination process
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
=
11
b
b
d
d
01l
00111
* dlbd
bd
−=
=
Forward
(1) Doolittle Decomposition (Factorization): “cont.”
© Sameh Nassar
=
3
2
3
2
b
b
d
d
1ll
01l
3231
21
1233
122
**
*
dldlbd
dlbd
3132
21
−−=
−= Forward Substitution
=
3
2
1
3
2
1
d
d
d
x
x
x
a"00
a'a'0
aaa
33
2322
131211
111213
2223
33
axaxadx
a'xa'dx
a"dx
/)**(
/)*(
/
2311
322
33
−−=−=
=Back
Substitution
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
(1) Doolittle Decomposition (Factorization): “cont.”
All l ii are equal to 1.
© Sameh Nassar
Note that pivoting is required if any uii is zero
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
Forward nifordlbdi
,....,3,21
=−= ∑−
(1) Doolittle Decomposition (Factorization): “cont.”
bd =
© Sameh Nassar
Forward Substitution
Back Substitution1,....,2,11 −−=
−=
∑+= nnifor
u
xud
xii
n
ijjiji
inn
nn u
dx =
nifordlbdj
jijii ,....,3,21
=−= ∑=
11 bd =
Note that pivoting is required if any uii is zero
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #4
Solve the following system of equations by LU
decomposition (Doolittle approach) without pivoting:
114 =−+ xx8x
© Sameh Nassar
762
114
=+−
=++−
=−+
321
321
321
xxx
4x5x2x
xx8x
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #4 (Sol. “1/3”)
762
114
=+−=++−
=−+
321
321
321
xxx
4x5x2x
xx8x
=7
4
11
B
−=
612
152-
1-48
A
© Sameh Nassar
1332211 === lll
25.08
2
11
2121 −=−==
u
al81111 == au
41212 == au
11313 −== au25.0
8
2
11
3131 ===
u
al
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #4 (Sol. “2/3”)
64*)25.0(5* 12212222 =−−=−= ulau
75.01*)25.0(1* 13212323 =−−−=−= ulau
33333.06
4*25.01* 12313232 −=−−=−=
u
ulal
© Sameh Nassar
5.600
0.7560
1-48
UL
−−
133333.025.0
25.0 01
001
622u
5.675.0*)33333.0(1*25.06** 233213313333 =−−−−=−−= ululau
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #4 (Sol. “3/3”)
=
−− 4
11
25.0 2
1
d
d
01
001
5.611*25.75.6*(7
75.611*)25.04
11
2
1
=−−==−−=
=
0-0.33333)d
(d
d
© Sameh Nassar
− 7133333.025.0 3d 5.611*25.75.6*(73 =−−= 0-0.33333)d
=
5.6
75.6
11
5.600 3
2
1
x
x
x
0.7560
1-48
1.0/)0.1*0.1*)1(11(
0.1/)0.1*75.75.6(
0.15./5.6
1
2
3
=−−−==−=
==
84x
60x
6x
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
(2) Crout’s Method:
All uii are
equal to 1.
© Sameh Nassar
Note that pivoting is required if any lii is zero
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
LU Decomposition
Forward dlbi
jiji
1
−∑−
(2) Crout’s Method: “cont.”
1bd =
© Sameh Nassar
Forward Substitution
Back Substitution
1,....,2,11
−−=−= ∑+=
nniforxudxn
ijjijiinn dx =
niforl
dlb
dii
jjiji
i ,....,3,21 =−
=∑
=11
11 l
bd =
Note that pivoting is required if any lii is zero
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #5
Solve the following system of equations by LU
decomposition (Crout’s approach) without pivoting:
114 =−+ xx8x
© Sameh Nassar
762
114
=+−
=++−
=−+
321
321
321
xxx
4x5x2x
xx8x
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #5 (Sol. “1/3”)
762
114
=+−=++−
=−+
321
321
321
xxx
4x5x2x
xx8x
=7
4
11
B
−=
612
152-
1-48
A
© Sameh Nassar
1332211 === uuu
5.08
4
11
1212 ===
l
au81111 == al
22121 −== al
23131 == al125.0
8
1
11
1313 −=−==
l
au
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #5 (Sol. “2/3”)
65.0*)2(5* 12212222 =−−=−= ulal
125.06
125.0*)2(1*
22
13212323 =−−−=−=
l
ulau
25.0*21* −=−−=−= ulal
© Sameh Nassar
1000.125100.125-0.51
UL
−−
5.6222 06
008
5.6125.0*)2(125.0*26** 233213313333 =−−−−=−−= ululal
25.0*21* 12313232 −=−−=−= ulal
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #5 (Sol. “3/3”)
=
−−
7411
5.6222 2
1
ddd
06008
15.6/]125.1*)2(375.1*27[
125.16/]375.1*)24[
375.18/11
2
1
=−−−==−−=
==
d
(d
d
© Sameh Nassar
− 75.622 3d 15.6/]125.1*)2(375.1*27[3 =−−−=d
=
1125.1375.1
100 3
2
1
xxx
0.125100.125-0.51
1.01*50.01*)125.0(375.1
0.1)0.1*125.125.1(
0.1
1
2
3
=−−−==−=
=
x
0x
x
Same results as Doolittle (Ex. #4)
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
The Matrix Inverse
Why the Matrix Inverse is needed for solving systems of linear equations?
bA.x = .bAx -1=
Note that the condition for the existence of A-1 is that: |A| # 0.
© Sameh Nassar
=
100010001
zzzzzzzzz
aaaaaaaaa
333231
232221
131211
333231
232221
131211
Case of 3 Equations (for example), i.e. n =3:
-1AA I
By definition: where I is the identity matrix
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Matrix Inverse using LU Decomposition
Using LU Decomposition
(where M is an intermediate Matrix)
The matrix inverse can be obtained using LU Decomposition
© Sameh Nassar
=−
333231
232221
131211
zzzzzzzzz
1A
Case of 3 Equations (for example), i.e. n =3:
=
333231
232221
131211
mmm
mmm
mmm
M
31
21
11
m
m
mThe vectors: , and will be obtained as the vector.
3
2
1
d
d
d
32
22
12
m
m
m
33
23
13
m
m
m
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
=
=
333231
232221
131211
333231
232221
131211
3231
21
III
III
III
100
010
001
mmm
mmm
mmm
1ll
01l
001
Doolittle LU Decomposition:
Matrix Inverse using LU Decomposition
© Sameh Nassar
3332313332313231 III100mmm1ll
LM = I = I
Forward Substitution Compute M
11223232
122222
1212
**
*
mlmlIm
mlIm
Im
3132
21
−−=
−=
=
13233333
132323
1313
**
*
mlmlIm
mlIm
Im
3132
21
−−=
−=
=
11213131
112121
1111
**
*
mlmlIm
mlIm
Im
3132
21
−−=
−=
=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
=
232221
131211
232221
131211
2322
131211
mmm
mmm
mmm
zzz
zzz
zzz
a"00
a'a'0
aaa
Matrix Inverse using LU Decomposition
Doolittle LU Decomposition:
© Sameh Nassar
33323133323133 mmmzzza"00
UA-1 = M
Back Substitution Compute A-1
111213
2223
33
azazamz
a'za'mz
a"mz
/)**(
/)*(
/
21311111
312121
3131
−−=
−=
=
111213
2223
33
azazamz
a'za'mz
a"mz
/)**(
/)*(
/
22321212
322222
3232
−−=
−=
=
111213
2223
33
azazamz
a'za'mz
a"mz
/)**(
/)*(
/
23331313
332323
3333
−−=
−=
=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #6
114 =−+ 321 xx8x
Solve the following system of equations using the
matrix inverse approach (LU by Doolittle approach):
© Sameh Nassar
762 =+−
=++−
321
321
xxx
4x5x2x
- Use the L and U matrices obtained from Example #4.
- When obtaining A-1 , check it by verifying [A][A-1] = [I].
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #6 (Sol. “1/5”)
−=
612
152-
1-48
A
From Example #4
=7
4
11
b
© Sameh Nassar
=5.600
0.7560
1-48
U
−−=
133333.025.0
25.0 01
001
L
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #6 (Sol. “2/5”)
=
−−
100
010
001
mmm
mmm
mmm
01
001
333231
232221
131211
133333.025.0
25.0
LM = I
© Sameh Nassar
33333.00*25.1*)33333.0(0
10*)25.0(1
0
32
22
12
=−−−==−=
=
0m
-m
m
10*25.0*)33333.0(1
00*)25.00
0
33
23
13
=−−−==−−=
=
0m
(m
m
-0.166671*25.25.0*)33333.00
25.01*)25.0(0
1
31
21
11
=−−−==−−=
=
0(m
m
m
−
=
133333.016667.0
0125.0
001
M
LM = I
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #6 (Sol. “3/5”)
−=
133333.016667.0
25.0
5.600
01
001
zzz
zzz
zzz
0.7560
1-48
333231
232221
131211
UA-1 = M
© Sameh Nassar
UA-1 = M
93590.090.04487)/8*4-0.02564*1)(1(
0.0448726-0.02564)/*0.750.25(
-0.02564150.16667/6.
11
21
31
=−−−==−=
=−=
z
z
z
.=−−−==−=
==
073718-080.160256)/*40.051282*1)(0(
0.16025660.051282)/*0.75(1
0.05128250.33333/6.
12
22
32
z
z
z
0.028846)/801923-0*40.153846*1)((0
019231-060.153846)/*0.75(0
0.1538461/6.5
13
23
33
= .−−−= .=−=
==
z
z
z
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #6 (Sol. “4/5”)
−−
−=−
153846.0051282.0025641.0
019231.0160256.0044872.0
028846.0073718.0099359.01A
© Sameh Nassar
Check:
=
−−
−
−=−
100
010
001
152-
1-48
153846.0051282.0025641.0
019231.0160256.0044872.0
028846.0073718.0099359.0
612
1AA
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #6 (Sol. “5/5”)
.bAx -1=
=7
4
11
b
−−
−=−
153846.0051282.0025641.0
019231.0160256.0044872.0
028846.0073718.0099359.01A
− 11028846.0073718.0099359.0x
=
3
2
1
x
x
x
x
© Sameh Nassar
−−
−=
=7
4
11
.
153846.0051282.0025641.0
019231.0160256.0044872.0
028846.0073718.0099359.0
3
2
1
x
x
x
x
=
++−−++−
=1
1
1
7*153846.04*051282.011*025641.0
7*019231.04*160256.011*044872.0
7*028846.04*073718.011*099359.0
Same results as Ex. #4
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Matrix Inverse using Gauss-Jordan Elimination
=
010
001
zzz
zzz
aaa
aaa
232221
131211
232221
131211
Case of 3 Equations (for example), i.e. n =3:
By definition: where I is the identity matrix
© Sameh Nassar
100zzzaaa 333231333231
-1AA I
=
0
0
1
31
21
11
z
z
z
aaa
aaa
aaa
333231
232221
131211
=
0
1
0
32
22
12
z
z
z
aaa
aaa
aaa
333231
232221
131211
=
1
0
0
33
23
13
z
z
z
aaa
aaa
aaa
333231
232221
131211
A x1 = b1 A x2 = b2 A x3 = b3
Hence, Gauss-Jordan elimination (for example) can be used to solve 3 systems of linear equations of the forms (i.e. applying it 3 times in this case):
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Recall: Solving Systems of Linear Equations (SLEs)
Methods:
(1) Gauss Elimination
(2) Gauss-Jordan Elimination Direct (Elimination) Methods
© Sameh Nassar
(2) Gauss-Jordan Elimination
(3) LU Decomposition
(4) Jacobi
(4) Gauss-Seidel
Direct (Elimination) Methods
Iterative methods
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Iterative Methods for Solving SLEs
• Called iterative methods, and hence, are similar in spirit to the iterative techniques of root finding (Ch. 3), specifically the fixed-point iteration technique.
• Employ initial guesses and then iterates to obtain refined
© Sameh Nassar
• Employ initial guesses and then iterates to obtain refined estimates of the solution.
• These methods are particularly suitable for large number of equations where elimination methods can be subject to round-off errors.
• However, there are certain instances where the method will not converge to the correct answer.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Iterative Methods for Solving SLEs
Case of 3 Equations (for example), i.e. n =3:
11
1312
a
xaxabx 321
1
−−=
© Sameh Nassar
11a
22
2321
a
xaxabx 312
2
−−=
33
3231
a
xaxabx 213
3
−−=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Iterative Methods for Solving SLEs
(1) Gauss-Seidel Method:
© Sameh Nassar
(2) Jacobi Method:
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Gauss-Seidel Method
%100x
xx||
j,i
1j,ij,ijai
−−=ε
where:
i = 1, 2,…., n
1st
Iteration
© Sameh Nassar
i = 1, 2,…., n
j = 1, 2,…. is the
iteration number
2nd
Iteration
Note that the unknowns are updated sequentially in every iteration using the most updated value for each unknown.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Jacobi Method
%100x
xx||
j,i
1j,ij,ijai
−−=ε
where:
i = 1, 2,…., n
1st
Iteration
© Sameh Nassar
i = 1, 2,…., n
j = 1, 2,…. is the
iteration number
2nd
Iteration
Note that the unknowns are updated only once (i.e. all together) in every iteration using the values for all unknowns from the previous iteration.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Iterative Methods for Solving SLEs
∑=
>n
1jijii |a||a|
Convergence Criterion
= 3n333231
2n232221
1n131211
a......aaa
a......aaa
a......aaa
A
© Sameh Nassar
≠=
ij1j
nnn3n2n1 a......aaa
MMMMM
• Condition: the absolute value of the main diagonal element must be greater than the sum of the absolute values of the off-diagonal elements for each row.
• Such systems are called Diagonally Dominant.
• This condition is not necessary for convergence but it guarantees convergence.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #7
Use the Gauss-Seidel method to solve the following
system of equations to a tolerance of εs = 5%:
50x12x3x- =++
© Sameh Nassar
Start with initial values of zero for all unknowns.
40x9xx6
3xxx6
50x12x3x-
321
321
321
=++
=−−
=++
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #7 (Sol. “1/3”)
40x9xx6
3xxx6
50x12x3x-
321
321
321
=++
=−−
=++
50x12x3x-
40x9xx6
3xxx6
321
321
321
=++
=++
=−−
Diagonally Dominant
© Sameh Nassar
40x9xx6 321 =++
6
xxx 32
1
3 ++=
50x12x3x- 321 =++
0xxx000 321 ===
9
xx6x 31
2
40 −−=
12
xx3x 21
3
50 −+=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #7 (Sol. “2/3”)
3.9490744.111110.5*350
x
4.111119
00.5*640x
0.56
003x
1
1
2
1
=−+=
=−−=
=++=
%100100%03.949074
|ε|
100%100%4.11111
04.11111|ε|
100%100%0.5
00.5|ε|
1
1a
1a
2
1
=−=
=−=
=−=1st Iteration
© Sameh Nassar
3.94907412
4.111110.5*350x
13 =−+= %100100%3.949074
03.949074|ε| 1
a3=−=
2nd Iteration
4.39611212
2.7767491.843364*350x
2.7767499
3.9490741.843364*640x
1.8433646
3.9490744.111113x
2
2
2
3
2
1
=−+=
=−−=
=++=
10.17%100%4.396112
3.949074-4.396112|ε|
48.05%100%2.776749
4.11111-2.776749|ε|
72.88%100%1.843364
0.51.843364|ε|
2a
2a
2a
3
2
1
==
==
=−=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #7 (Sol. “3/3”)
iteration unknown value εεεεa maximum εεεεa x1 0.5 100.00%
1 x2 4.111111 100.00% x3 3.949074 100.00% 100.00%
x 1.843364 72.88% 72.88%
© Sameh Nassar
x1 1.843364 72.88% 72.88% 2 x2 2.776749 48.05% x3 4.396112 10.17%
x1 1.695477 8.72% 8.72% 3 x2 2.82567 1.73% x3 4.355063 0.94%
x1 1.696789 0.08% 4 x2 2.829356 0.13% 0.13% x3 4.355084 0.00%
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #8
Use the Jacobi method to solve the following
system of equations:
50x12x3x- 321 =++
© Sameh Nassar
Start with initial values of zero for all unknowns and
apply 2 iterations only.
40x9xx6
3xxx6
321
321
321
=++
=−−
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #8 (Sol. “1/2”)
40x9xx6
3xxx6
50x12x3x-
321
321
321
=++
=−−
=++
50x12x3x-
40x9xx6
3xxx6
321
321
321
=++
=++
=−−
Diagonally Dominant
© Sameh Nassar
40x9xx6 321 =++
6
xxx 32
1
3 ++=
50x12x3x- 321 =++
0xxx000 321 ===
9
xx6x 31
2
40 −−=
12
xx3x 21
3
50 −+=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #8 (Sol. “2/2”)
4.16666700*350
x
4.4444449
00*640x
0.56
003x
1
1
2
1
=−+=
=−−=
=++=
%100100%04.166667
|ε|
100%100%4.444444
04.444444|ε|
100%100%0.5
00.5|ε|
1
1a
1a
2
1
=−=
=−=
=−=1st Iteration
© Sameh Nassar
4.16666712
00*350x
13 =−+= %100100%4.166667
04.166667|ε| 1
a3=−=
2nd Iteration
3.92129612
4.4444440.5*350x
3.6481489
4.1666670.5*640x
1.9351856
4.1666674.4444443x
2
2
2
3
2
1
=−+=
=−−=
=++=
6.26%100%3.921296
4.166667-3.921296|ε|
%21.83100%3.648148
4.444444-3.648148|ε|
%74.16100%1.935185
0.51.935185|ε|
2a
2a
2a
3
2
1
==
==
=−=
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
True Errors and Residuals
• For a system of linear equations Ax = b with true solution vector xtrue solution and numerical “approximate” solution vector xnumerical solution
, the following error vectors are defined:
True Error Vector e: e = xtrue solution - xnumerical solution
© Sameh Nassar
Residual Vector r (or simply Residuals):
r = Axtrue solution - Axnumerical solution = b - Axnumerical solution
• Similar to what discussed before for single variables (Lecture #2), rcan be estimated while e is usually unknown.
• However, r does not really indicate how small the errors in the solution are and hence a small residual vector r does not guarantee a small true error vector e.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
True Error and Residual Vectors
• Assume we have:
© Sameh Nassar
• Assume the following 2 numerical solution vectors were obtained:
• Let’s now obtain the true error vectors and the corresponding residual vectors of the 2 cases (numerical solutions) and then compare the results.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
True Error and Residual Vectors
e is sensitive to the errors in the
numerical solution
© Sameh Nassar
r is not sensitive to the errors in the
numerical solution
• Note that in case (1), the residuals (r1) have the same order of magnitude as the true errors (e1), while in case (2) they are not.
• This is a typical example of an ill-conditioned coefficient matrix A.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
• Based on the discussions in previous slides, a more accurate estimate of the error in a numerical solution (of SLEs) should be obtained.
• In order to relate the magnitude of the residuals to the magnitude of the true errors, and also to determine the bounds for these
Norms
© Sameh Nassar
of the true errors, and also to determine the bounds for these errors, we are going to use quantities that measure the magnitude of matrices and vectors.
• These quantities are called norms.
• A norm is a non-negative real number assigned to a matrix or a vector.
• The norm of a vector νννν is denoted by ||ν||||ν||||ν||||ν||.
• The norm of a matrix A is denoted by ||||||||[A]|||||||| (or simply ||||||||A|||||||| ).
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Vector Norms
Categories (Types):
© Sameh Nassar
Properties:
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Matrix Norms
Categories (Types):
© Sameh Nassar
Where νννν is an eigenvector of the matrix A corresponding to an eigenvalue λ(discussed later!)
Euclidean
2-norm ||||||||[A] |||||||| =
(m x n matrix)
Additional Property:
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Error Bounds of SLEs Numerical Solutions
• For a system of linear equations Ax = b with true solution vector xT
and numerical “approximate” solution vector xN , the following terms (using norms) are defined:
Relative Error Re: Re = , e = xT - xN (true error)
© Sameh Nassar
Relative Residual Rr: Rr = , r = b - AxN (residual)
• Thus, a major objective here is to use Rr (that can be estimated from the numerical solution) to obtain “or estimate” an upper and lower bounds of Re (recall that Re itself is unknown).
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Error Bounds of SLEs Numerical Solutions
© Sameh Nassar
Lower Bound Upper Bound
• The term is called the Condition Number of A.
Lower Bound Upper Bound
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Condition Number of a Matrix
Properties:
(I is the identity matrix)
• If Cond(A) is large (i.e. >> 1), a small Rr will not guarantee a small
© Sameh Nassar
• If Cond(A) is large (i.e. >> 1), a small Rr will not guarantee a small Re (i.e. small changes in b or the elements of A may result in large
changes in x).
• In this case, the SLE (and hence A) is ill-conditioned.
• If Cond(A) is small (i.e. ~ 1), a small Rr will indicate a small Re (i.e.
small changes in b or the elements of A will result in small changes in x).
• In this case, the SLE (and hence A) is well-conditioned.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Eigenvalue Problems
• Known also as characteristic-value problems.
• An eigenvalue is a characteristic value of a square matrix (n x n).
Background #1:
If A is an n x n matrix, usually there is no obvious geometric
© Sameh Nassar
• If A is an n x n matrix, usually there is no obvious geometric relationship between a vector x and its image Ax (i.e. when A is multiplied by x).
• However, there are some nonzero vectors “x” that A is mapping into scalar multiples of themselves.
• These types of vectors arise in several engineering applications that involve vibrations, electrical systems, quantum mechanics, mechanical stress and elasticity.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Eigenvalue Problems
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No Geometric Relation between x and Ax
Ax is a scalar multiple of x
“x is an eigenvector”
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Eigenvalue Problems
• In the latter case, Ax = λλλλx, where λλλλ is a scalar.
• The scalar λλλλ is called an eigenvalue (or characteristic value) of Aand x is said to be an eigenvector corresponding to λλλλ.
Background #2:
(a) The system of equations: A.x = b
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(a) The system of equations: A.x = bis called non-homogeneous
(b) The system of equations: A.x = 0is called homogeneous
Why?
• For case (a), a unique solution of x can be obtained if the equations are linearly independent (i.e. |A| ≠ 0).
• For case (b), nontrivial solutions (any solutions rather than x = 0) can be obtained, however, they are generally not unique (i.e. several solutions of x are available).
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Eigenvalues λλλλ & Eigenvectors x
• The non-unique solution for x is obtained since several available relationships can be established between the x-vector elements that can be satisfied by various combinations of the x-values.
• To obtain the relations of the x-values, the eigenvalues λλλλ of A are obtained first, where the general form of the equations will be:
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A.x = λλλλ.x A.x - λλλλ.I.x = 0 [A - λλλλ.I].x = 0
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Eigenvalues λλλλ & Eigenvectors x
• Hence, the unknown parameter λλλλ (eigenvalue) should be obtained and then the relations for x-values can be evaluated.
Condition to obtain λλλλ for nontrivial solutions of x:
|A - λλλλ.I| = 0
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• The values of λλλλ can be obtained by expanding the determinant |A - λλλλ.I| and equalize to zero, which will yield a polynomial in the unknown λλλλ.
• The values of λλλλ are the roots of the polynomial (any method of solving SLEs can be used).
• Then, the eigenvector x can be evaluated for each value of λλλλ.
• For any constant c, cx is also an eigenvector of A (therefore, the eigenvectors x are normalized to be unit vectors, i.e. ||||||||x||||||||2= 1.
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #9
Find the eigenvalues of the given matrix:
−−
=23
A
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Then, evaluate the corresponding eigenvectors.
=01
A
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #9 (Sol. “1/3”)
−−=
01
23A
Eigenvalues:
|A - λλλλ.I| = 0 001
23=
−−−−
λλ
0)1*2(*)3( =−−−−− λλ 2 =++ λλ
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0)1*2(*)3( =−−−−− λλ 0232 =++ λλ
1*2
2*1*433 2 −±−=λ2
13±−=λ
11 −=λ 22 −=λ
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #9 (Sol. “2/3”)
Eigenvectors:
==
−
−−−00
001
23
2
1
xx
λλ
[A - λλλλ.I].x = 0
λ = λ1 = -1:
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0)1(01
2)1(3
2
1 =
−−−−−−
x
x0
1122
2
1 =
−−xx
12 xx −=
Use ||||||||x||||||||2= 1 122
21 =+ xx 1)( 2
121 =−+ xx
2
11 =x
2
12 −=x
−=
2
12
1
1λx
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
Example #9 (Sol. “3/3”)
Eigenvectors:
02)2(3 1 =
−−−− x
021 1 =
−− x 1
xx −=
λ = λ2 = -2:
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0)2(01 2
1 =
−− x
021
21
2
1 =
−−x
x12 2
1xx −=
Use ||||||||x||||||||2= 1 122
21 =+ xx 1)
2
1( 2
121 =−+ xx
5
21 =x
5
12 −=x
−=
5
15
2
2λx
ENGG 407 L20 Spring 2012 Lecture #3 Dr. Sameh Nassar
• Sections:
4.1
4.2
4.3
Textbook Readings
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4.3
4.4
4.5
4.6
4.7
4.10
4.11
4.12 (excluding 4.12.1 – 14.12.4)