mdb3053 chap9 linearalgebra part 1 may16
TRANSCRIPT
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RECAP ON CHAPTER 6
1. Bracketing Method:
Bisection Method
False-Position Method
2. Open Method:
Fixed Point Iteration
Newton-Raphson Method
Secant Method
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2x
1x
How to solve a system of
simultaneous algebraic
equations using CM ?
Cokes example
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LINEAR ALGEBRAIC EQUATIONS
Example of a system of L.A.E
Why coupled? Each equation has terms in common with the
others, that is x1, x2 and x3.
Why linear? It means the effects areproportionalto their cause.
Each equation contains only first order terms of x1, x2 and x3. No
terms like x12, or log(x3) or 1/x2x3.
132
132
8253
321
321
321
xxx
xxx
xxx
Three coupled,
linear equations
bxA
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GENERAL FORM OF L.A.E
The general form of a system of L.A.E forn number ofequations
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111coefficients
constants
unknown
variables
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MATRIX FORM OF L.A.E
Linear Algebraic Equations in matrix form
54
132
21
21
xx
xx
nnmnm
n
b
b
b
x
x
x
aa
aa
A
11
1
111
,,
bxA
m x n
coefficient
matrix, A
n x 1 column vector of
unknowns
n x 1 column vector
of constants
5
1
14
32
5
1
14
32
2
1
x
x
or
Example: augmented formbxA
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TRY THIS
Write the matrix representation for the L.A.Eshown below:
132
132
8253
321
321
321
xxx
xxx
xxx
1
1
8
321
132
253
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SPECIAL TYPES OF MATRICES
33
2322
131211
00
0
a
aa
aaa
333231
2221
11
0
00
aaa
aa
a
33
22
11
00
00
00
a
a
a
100
010
001
Diagonal matrix
Identity matrix, I
Lowertriangular matrix, L
Uppertriangular matrix, U
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WAYS TO SOLVE L.A.E
Graphical Methodby plottingCramers Ruleby determinant concept
Direct Method Gaussian Elimination(Exact Sol) Gauss-Jordan
LU Factorization
Iterative Method Jacobis Method
(Approx. Sol) Gauss-Seidel Method
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GRAPHICAL METHOD
The SOLUTION isshown graphically
They are in the forms of
straight linebecause of
linear equations
22
1823
21
21
xx
xx
)2(..........1
2
)1.......(2
39
12
12
xx
xx
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BASIC CONCEPTS OF SOLUTION
Unique Solution (Nonhomogeneous equation)
Without Solution
2x
1x
63
593
21
21
xx
xx
1x
2x
parallel lines
864
432
21
21
xx
xxLinearly
dependentequations
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Infinite # of Solutions (No unique solutions)
ill-conditioned slopes are close to each
other, very hard to determine the solution
BASIC CONCEPTS OF SOLUTION (CONT)
864
432
21
21
xx
xx
2x
1x
linearly
dependant
5.522
7.52.22
21
21
xx
xx
1x
2x
identical lines
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GAUSSIAN ELIMINATION
Gaussian = Forward + Back
Elimination Elimination Substitution
To transform the original
matrix, A into an upper, U
triangular matrix
Unknowns are solved
through Back Substitution
in the upper triangular
matrix
Most frequently used Direct Method
Principle: to reduce coefficient matrix, A into
equivalent upper triangular form
Consists ofTWO stages:
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Fig. 9.3
upper
triangular
matrix
Augmented
form
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FORWARD ELIMINATION
Forward Elimination stage reduces the augmented
matrix into an uppertriangular matrix.
It requires Elementary Row Transformations
(3 Basic approaches). You can:
1. Interchange any two row in position
2. Multiply (or divide) a row by a
nonzero scalar , k=2 (eg. 2 R2)
3. Multiply (or divide) a row by ascalar and adding it (or subtracting
it from) another row (eg R1 + 2 R2)
1
5
32
14
5
1
14
32
2
5
64
14
7
5
70
14
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Example: Reducing matrix to upper matrix
1 2 2 3
Consider the matrix 4 4 2 6
4 6 4 10
Lets perform the transformation at second row R2:
new R2
It means To get the second row of the new matrix,
multiply every element of the first row by 4 and subtract
it from the element in the second row.
12
'
2
1
4RRR
R1
R2
R3
1
4 Elimination factor or
multiplier
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Thus, the new 2nd row becomes :
R2 = [4 4 2 | 6] 4 [1 2 2 | 3]
= [0 4 6 |6]
EXAMPLE (cont)
1221
4RRR
133
1
4RRR
You can perform one
more transformation to
obtain U matrix
Original Matrix:
1 2 2 3
4 4 2 6
4 6 4 10
Transformation:
1 2 2 3
0 -4 -6 -6
0 -2 -4 -2
Next, transformation at 3rd row
Thus, the new 3rd row becomes:
R3 = [4 6 4 | 10] 4 [1 2 2 | 3]
= [ 0 2 4 |2]R
3
'' = [0 -2 -4 | -2] 1/2 [0 -4 -6 | -6]
= [ 0 0 1 | 1]
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BACK SUBSTITUTION
Suppose that the next forward elimination step results in
the following augmented matrix:
1 2 2 3
0 4 6 6
0 0 1 1
All the entries
below the
diagonal are
ZEROS
By using Back Substitution, SOLUTIONS are:
R3: x3 = 1 x3 =1
R2: 4x26x3 =6 x2 = 3
R1 : x1 + 2x2 + 2x3 = 3 x1 =1
Direct method
yields exact sol
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CLASS ACTIVITY #1
Use fraction number to reduce round-off errors.
3232
14
232
321
321
321
xxx
xxx
xxx
Try this in EXCEL: (p296)
To inverse, highlight the
range, enter minverse(..)
and press Ctrl+Shift+Enter
To multiply,enter mmult(..)
Try this in MATLAB: (p298)
>> A=[2 1 3;4 1 1;-2 3 2]
>> b=[2 ; 1 ; 3 ]
>> x = A\b
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CLASS ACTIVITY #1 (cont)
Step 1: Augmented Matrix Form:
3232
14
232
321
321
321
xxx
xxx
xxx
3
1
2
232
114
312
462423
5
2
232
530
312
2 112
RRR
Step 2: Forward Elimination by performing
elementary row transformation
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Pitfalls of Gaussian Elimination
Division by zero. It is possible that during bothelimination and back-substitution phases a division by
zero can occur. can be solved by (Partial) Pivoting
Round-off errors. Because computer can only store afixed no. of digits to use more S.F. orfractions
Ill-conditioned systems. Systems where small changes
in coefficients result in large changes in the solution. Ithappens when two or more equations are nearly
identical.
1x
2x
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GAUSS-JORDAN
A variation of Gauss elimination but 50% more
computational work than Gaussian Elimination
Elimination step results in an identity matrix (not
upper triangular matrix) due to normalization step
No back substitution required
3
2
1
332331
232221
131211
c
c
c
aaa
aaa
aaa
3
2
1
100
010
001
h
h
h
Elimination +Normalization
Identity
matrix
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Fig 9.9
IdentityMatrix
No back
substitution !
Elimination +
Normalization
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EXAMPLE
Suppose that the forward elimination and normalizing
steps result in the following augmented matrix:
1 0 0 5
0 1 0 3
0 0 1 7
The SOLUTION will be : x1 = 5 x2 = 3 and x3 = 7
Hence there is no need for back substitution
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CLASS ACTIVITY #2
Solve the system of equations below using
Gauss-Jordan method
123
6252
72
321
321
321
xxx
xxx
xxx
Ans: x1 = 2 ; x2 = 8; x3 = 21
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MATRIX INVERSION
Inverse Matrix A-1
can be found through Gauss-Jordan Method
STEP 1 : Form
STEP 2 : Transform to using theElementary Row Transformations.
STEP 3 : Finally, B = A -1, where B is the inversematrix
IA
IA BI
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CLASS ACTIVITY #3
MATRIX INVERSION
Find the inverse of the matrix below usingthe Gauss-Jordan Method
612
131
221
A
135
124
410171
A