ecuaciones algebraicas lineales-1
Post on 04-Apr-2018
224 Views
Preview:
TRANSCRIPT
-
7/29/2019 Ecuaciones Algebraicas lineales-1
1/59
Ecuaciones Algebraicas lineales
-
7/29/2019 Ecuaciones Algebraicas lineales-1
2/59
An equation of the form ax+by+c=0 or equivalently ax+by=-
c is called a linear equation inx andy variables.
ax+by+cz=dis a linear equation in three variables,x, y, and
z.
Thus, a linear equation in n variables is
a1x1+a2x2+ +a
nxn = b A solution of such an equation consists of real numbers c1, c2,
c3, , cn. If you need to work more than one linear
equations, a system of linear equations must be solved
simultaneously.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
3/59
Matricesaij = elementos de una matriz
i=nmero del rengln
j=nmero de la columna
Vector rengln
Vector columna
-
7/29/2019 Ecuaciones Algebraicas lineales-1
4/59
Matriz cuadrada m=n
Diagonal principal
Nmero de incngnitas
Nmero de
ecuaciones
-
7/29/2019 Ecuaciones Algebraicas lineales-1
5/59
Reglas de operaciones con matrices
-
7/29/2019 Ecuaciones Algebraicas lineales-1
6/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
7/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
8/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
9/59
Representacin de ecuaciones algebraicas
lineales en forma matricial
Solving for X
-
7/29/2019 Ecuaciones Algebraicas lineales-1
10/59
Part 3 10
Noncomputer Methods for Solving
Systems of Equations
For small number of equations (n 3) linear
equations can be solved readily by simple
techniques such as method of elimination.
Linear algebra provides the tools to solve such
systems of linear equations.
Nowadays, easy access to computers makes
the solution of large sets of linear algebraic
equations possible and practical.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
11/59
Part 3 11
Gauss EliminationChapter 9
Solving Small Numbers of Equations
There are many ways to solve a system of
linear equations:
Graphical method
Cramers rule
Method of elimination
Computer methods
For n 3
-
7/29/2019 Ecuaciones Algebraicas lineales-1
12/59
Part 3 12
Graphical Method
For two equations:
Solve both equations for x2:
2222121
1212111
bxaxa
bxaxa
22
21
22
212
1212
1
112
11
2 intercept(slope)
a
bx
a
ax
xxa
bx
a
ax
-
7/29/2019 Ecuaciones Algebraicas lineales-1
13/59
Part 3 13
Plot x2 vs. x1
on rectilinear
paper, the
intersection of
the lines
present the
solution.
Fig. 9.1
-
7/29/2019 Ecuaciones Algebraicas lineales-1
14/59
Part 3 14
Graphical Method
Or equate and solve for x1
12
11
22
21
12
1
22
2
12
11
22
21
22
2
12
1
1
22
2
12
11
12
11
22
21
22
21
22
21
12
11
12
112
0
a
a
a
a
a
b
a
b
a
a
a
a
a
b
a
b
x
a
b
a
bx
a
a
a
a
a
bx
a
a
a
bx
a
ax
-
7/29/2019 Ecuaciones Algebraicas lineales-1
15/59
Part 3 15
Figure 9.2
No solution Infinite solutions Ill-conditioned
(Slopes are too close)
-
7/29/2019 Ecuaciones Algebraicas lineales-1
16/59
Part 3 16
Determinants and Cramers Rule
Determinant can be illustrated for a set of three
equations:
Where A is the coefficient matrix:
bAx
333231
232221
131211
aaa
aaa
aaa
A
-
7/29/2019 Ecuaciones Algebraicas lineales-1
17/59
Part 3 17
Assuming all matrices are square matrices, thereis a number associated with each square matrix A
called the determinant, D, of A. (D=det (A)). If[A] is order 1, then [A] has one element:
A=[a11]
D=a11 For a square matrix of order 2, A=
the determinant is D= a11 a22-a21 a12a11 a12
a21 a22
-
7/29/2019 Ecuaciones Algebraicas lineales-1
18/59
Part 3 18
For a square matrix of order 3, the minor of
an element aij is the determinant of the matrixof order 2 by deleting row i and columnj of A.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
19/59
Part 3 19
22313221
3231
2221
13
23313321
3331
2321
12
23323322
3332
2322
11
333231
232221
131211
aaaaaa
aaD
aaaaaa
aaD
aaaaaa
aa
D
aaa
aaa
aaa
D
-
7/29/2019 Ecuaciones Algebraicas lineales-1
20/59
Part 3 20
3231
2221
13
3331
2321
12
3332
2322
11aa
aaa
aa
aaa
aa
aaaD
Cramers rule expresses the solution of a
systems of linear equations in terms of ratios
of determinants of the array of coefficients ofthe equations. For example, x1 would be
computed as:
D
aab
aab
aab
x33323
23222
13121
1
-
7/29/2019 Ecuaciones Algebraicas lineales-1
21/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
22/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
23/59
Part 3 23
Method of Elimination
The basic strategy is to successively solve one
of the equations of the set for one of the
unknowns and to eliminate that variable from
the remaining equations by substitution.
The elimination of unknowns can be extended
to systems with more than two or three
equations; however, the method becomesextremely tedious to solve by hand.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
24/59
Relacin con Cramer
-
7/29/2019 Ecuaciones Algebraicas lineales-1
25/59
Part 3 25
Naive Gauss Elimination
Extension ofmethod of elimination to largesets of equations by developing a systematicscheme or algorithm to eliminate unknowns
and to back substitute. As in the case of the solution of two equations,
the technique for n equations consists of twophases:
Forward elimination of unknowns
Back substitution
-
7/29/2019 Ecuaciones Algebraicas lineales-1
26/59
Part 3 26
Fig. 9.3
-
7/29/2019 Ecuaciones Algebraicas lineales-1
27/59
Generalizando
Multiplicando ec 1
Restando ec2 de la nueva ec1
Reescribiendo ec anterior
a32/a22 = nuevo
elemento pivote
Elemento pivote
-
7/29/2019 Ecuaciones Algebraicas lineales-1
28/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
29/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
30/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
31/59
Part 3 31
Pitfalls of Elimination Methods
Division by zero. It is possible that during bothelimination and back-substitution phases a divisionby zero can occur.
Round-off errors.
Ill-conditioned systems. Systems where small changesin coefficients result in large changes in the solution.Alternatively, it happens when two or more equationsare nearly identical, resulting a wide ranges of
answers to approximately satisfy the equations. Sinceround off errors can induce small changes in thecoefficients, these changes can lead to large solutionerrors.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
32/59
Part 3 32
Singular systems. When two equations are
identical, we would loose one degree offreedom and be dealing with the impossiblecase ofn-1 equations for n unknowns. Forlarge sets of equations, it may not be obvious
however. The fact that the determinant of asingular system is zero can be used and testedby computer algorithm after the eliminationstage. If a zero diagonal element is created,calculation is terminated.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
33/59
Part 3 33
Techniques for Improving Solutions
Use of more significant figures.
Pivoting. If a pivot element is zero,normalization step leads to division by zero.
The same problem may arise, when the pivotelement is close to zero. Problem can beavoided: Partial pivoting. Switching the rows so that the
largest element is the pivot element.
Complete pivoting. Searching for the largestelement in all rows and columns then switching.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
34/59
Cramer o sustitucin
-
7/29/2019 Ecuaciones Algebraicas lineales-1
35/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
36/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
37/59
Determinant Evaluation Using Gauss Elimination
-
7/29/2019 Ecuaciones Algebraicas lineales-1
38/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
39/59
Casi cero !!!
Depende del numero de cifras
significativas
-
7/29/2019 Ecuaciones Algebraicas lineales-1
40/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
41/59
SCALING
-
7/29/2019 Ecuaciones Algebraicas lineales-1
42/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
43/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
44/59
Part 3 44
Gauss-Jordan
It is a variation of Gauss elimination. Themajor differences are:
When an unknown is eliminated, it is eliminated
from all other equations rather than just thesubsequent ones.
All rows are normalized by dividing them by theirpivot elements.
Elimination step results in an identity matrix. Consequently, it is not necessary to employ back
substitution to obtain solution.
-
7/29/2019 Ecuaciones Algebraicas lineales-1
45/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
46/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
47/59
Descomposicin LU e inversin de
-
7/29/2019 Ecuaciones Algebraicas lineales-1
48/59
Descomposicin LU e inversin de
Matrices
[A]{X}={B}
[A]{X}-{B}=0
[U]{X}-{D}=0
Gauss
Elimination
-
7/29/2019 Ecuaciones Algebraicas lineales-1
49/59
De la eliminacin hacia delante de Gauss
tenemos :
Finalmente
-
7/29/2019 Ecuaciones Algebraicas lineales-1
50/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
51/59
Encontrando d aplicando la eliminacin
hacia adelante pero solo sobre el vector B
Encontrando X aplicando la sustitucin
hacia atrs
-
7/29/2019 Ecuaciones Algebraicas lineales-1
52/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
53/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
54/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
55/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
56/59
Matriz Inversa
-
7/29/2019 Ecuaciones Algebraicas lineales-1
57/59
-
7/29/2019 Ecuaciones Algebraicas lineales-1
58/59
Homework
-
7/29/2019 Ecuaciones Algebraicas lineales-1
59/59
Homework
top related