[PDF] • Linear system with m equations and n unknowns/variables: a11x1





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• Linear system with m equations and n unknowns/variables: a11x1

If solution is unique there must be n pivots for an m × n system. • Reduced row echelon form is unique – can be used to show two linear systems are equivalent 



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Calcul symbolique et propagation des singularités pour les

une équation non linéaire d'ordre m où F est réelle et de classe C°°. Si uest une solution réelle de (0.1 ) on définit le symbole principal :.

Linear system withmequations andnunknowns/variables: a

11x1+a12x2+a13x3++a1nxn=b1

a

21x1+a22x2+a23x3++a2nxn=b2

a

31x1+a32x2+a33x3++a3nxn=b3

a m1x1+am2x2+am3x3++amnxn=bm Solve by "simplifying" the system by using elementary row operations that preserve solution set. Only need to keep track of changes toa's andb{ use matrices to store these numbers:2 6

6666664a

11a12a13a1nb

1 a

21a22a23a2nb

2 a

31a32a33a3nb

3.............

a m1am2am3amnb m3 7

7777775.

Systematically simplies torow echelon formor further toreduced row echelon form.

For example, the coecient matrix looks like :

row echelon form: 2 6

6664

0 0 0 0

0 0 0 0 03

7

7775; reduced row echelon:2

6

66641 00

0 10

0 0 0 1

0 0 0 0 03

7 7775
Row echelon form is sucient to conclude about existence and unique- ness of solutions: If system is consistent, row echelon form cannot have any rows with h 0 00b i whereb6= 0. If solution is unique, there must benpivots for anmnsystem. Reduced row echelon form is unique { can be used to show two linear systems areequivalentor the corresponding matrix representations arerow equivalent 1

Example to illustrate systematic row reduction:

2 6

40 0 123

17 0 6 5

1 64 2 73

7 5!2 6

417 0 6 5

0 0 123

1 64 2 73

7 5 2 6

417 0 6 5

0 0 123

014 8 123

7 5!2 6

417 0 6 5

014 8 12

0 0 1233

7 5 2 6

417 0 6 5

01 0 0 0

0 0 1233

7 5!2 6

417 0 6 5

0 1 0 0 0

0 0 1233

7 5 2 6

41 0 0 6 5

0 1 0 0 0

0 0 1233

7 5 If this is an augmented matrix that corresponds to a linear system: 2 6

41 0 0 65

0 1 0 00

0 0 1233

7 5=)x

1+6x4= 5

x 2= 0 x

32x4=3

x

1;x2;x3corresponds to thepivot positionsand are thebasic vari-

ables. The non-basic variablex4is called thefree variable, and is con- ventionally treated as a parameter. The general solution orparametric descriptionof the solution set is

8>>>>>><

>>>>>:x

1= 56x4

x 2= 0 x

3=3 + 2x4

x

4is freeor8

>>>>>:x

1= 56k

x 2= 0 x

3=3 + 2k

x

4=kwherekis the parameter:

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