Eliminação de Gauss

Eliminacao de Gauss

Loıc Cerf

13 de fevereiro de 2014UFMG – ICEx – DCC

Exercıcio A

Exercıcio

Resolver o sistema linear pelo metodo de eliminacao deGauss com pivotacao parcial, usando 4 casas decimais:−x1 + x2 − 2x3 = −20

5x1 + 2x2 + x3 = 21

2x1 + 5x2 + 4x3 = 33

.

Calcular o vetor resıduo e a norma-2 dele.

2 / 9Loıc Cerf Eliminacao de Gauss

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Exercıcio A

O sistema na forma matricial:

Ax = b com A =

−1 1 −25 2 12 5 4

, x =

x1

x2

x3

e b =

−202133

.


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − × L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − × L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0

1, 4 − 1, 8 − 15, 8

− −15 × L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − × L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4

− 1, 8 − 15, 8

− (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − × L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8

− 15, 8

− (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − × L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − × L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0

4, 2 3, 6 − 24, 6

− 25 × L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2

3, 6 − 24, 6

− 0, 4× L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2 3, 6

− 24, 6

− 0, 4× L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − 0, 4× L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − 0, 4× L2 + L3

L6 0 0

− 2, 9999 − 23, 9992

− 1,44,2 × L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − 0, 4× L2 + L3

L6 0 0 − 2, 9999

− 23, 9992

− 0, 3333× L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − 0, 4× L2 + L3

L6 0 0 − 2, 9999 − 23, 9992 − 0, 3333× L5 + L4


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Exercıcio A

A b operacoes

L1 − 1 1 − 2 − 20L2 5 2 1 21L3 2 5 4 33

L4 0 1, 4 − 1, 8 − 15, 8 − (−0, 2)× L2 + L1

L5 0 4, 2 3, 6 − 24, 6 − 0, 4× L2 + L3

L6 0 0 −2, 9999 − 23, 9992 − 0, 3333× L5 + L4


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Exercıcio A

Ax = b ∼

5 2 10 4, 2 3, 60 0 − 2, 9999

x1

x2

x3

=

2124, 6

− 23, 9992

.

Solucoes do sistema triangular superior obtidas pelas substituicoesretroativas:

x3 = −23,9992−2,9999

x2 = 24,6−3,6×84,2

x1 = 21−2×(−1)−1×85


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Exercıcio A

Ax = b ∼

5 2 10 4, 2 3, 60 0 − 2, 9999

x1

x2

x3

=

2124, 6

− 23, 9992

.

Solucoes do sistema triangular superior obtidas pelas substituicoesretroativas:x3 = −23,9992

−2,9999

x2 = 24,6−3,6×84,2

x1 = 21−2×(−1)−1×85


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Exercıcio A

Ax = b ∼

5 2 10 4, 2 3, 60 0 − 2, 9999

x1

x2

8

=

2124, 6

− 23, 9992

.

Solucoes do sistema triangular superior obtidas pelas substituicoesretroativas:x3 = 8

x2 = 24,6−3,6×84,2

x1 = 21−2×(−1)−1×85


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Exercıcio A

Ax = b ∼

5 2 10 4, 2 3, 60 0 − 2, 9999

x1

x2

8

=

2124, 6

− 23, 9992

.

Solucoes do sistema triangular superior obtidas pelas substituicoesretroativas:x3 = 8x2 = 24,6−3,6×8

4,2

x1 = 21−2×(−1)−1×85


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Exercıcio A

Ax = b ∼

5 2 10 4, 2 3, 60 0 − 2, 9999

x1

− 18

=

2124, 6

− 23, 9992

.

Solucoes do sistema triangular superior obtidas pelas substituicoesretroativas:x3 = 8x2 = − 1

x1 = 21−2×(−1)−1×85


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Exercıcio A

Ax = b ∼

5 2 10 4, 2 3, 60 0 − 2, 9999

x1

− 18

=

2124, 6

− 23, 9992

.

Solucoes do sistema triangular superior obtidas pelas substituicoesretroativas:x3 = 8x2 = − 1x1 = 21−2×(−1)−1×8

5


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Exercıcio A

Ax = b ∼

5 2 10 4, 2 3, 60 0 − 2, 9999

3− 18

=

2124, 6

− 23, 9992

.

Solucoes do sistema triangular superior obtidas pelas substituicoesretroativas:x3 = 8x2 = − 1x1 = 3


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Exercıcio A

Resultado −1 1 −25 2 12 5 4

x =

−202133

⇔ x =

3−18

.

O vetor resıduo e: −202133

− − 1 1 − 2

5 2 12 5 4

3− 18

=

−202133

−

− 202133

A norma-2 do vetor resıduo e√

02 + 02 + 02 = 0.


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Exercıcio A

Resultado −1 1 −25 2 12 5 4

x =

−202133

⇔ x =

3−18

.


− − 1 1 − 2

5 2 12 5 4

3− 18

=

−202133

− − 20

2133


02 + 02 + 02 = 0.


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Exercıcio A

Resultado −1 1 −25 2 12 5 4

x =

−202133

⇔ x =

3−18

.


− − 1 1 − 2

5 2 12 5 4

3− 18

=

−202133

− − 20

21

33


02 + 02 + 02 = 0.


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Exercıcio A

Resultado −1 1 −25 2 12 5 4

x =

−202133

⇔ x =

3−18

.


− − 1 1 − 2

5 2 12 5 4

3− 18

=

−202133

− − 20

2133


02 + 02 + 02 = 0.


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Exercıcio A

Resultado −1 1 −25 2 12 5 4

x =

−202133

⇔ x =

3−18

.


− − 1 1 − 2

5 2 12 5 4

3− 18

=

000


02 + 02 + 02 = 0.


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Exercıcio A

Resultado −1 1 −25 2 12 5 4

x =

−202133

⇔ x =

3−18

.


− − 1 1 − 2

5 2 12 5 4

3− 18

=

000

A norma-2 do vetor resıduo e

√02 + 02 + 02 = 0.


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Exercıcio B

Exercıcio

Resolver o sistema linear pelo metodo de eliminacao deGauss com pivotacao parcial, usando 2 casas decimais:

2x1 − x2 + 3x3 = 9

−x1 + 2x2 − x3 = 0

3x1 + 2x2 − 2x3 = 1

.

Calcular o vetor resıduo e a norma-1 dele.


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Exercıcio C

Exercıcio para entregar

Resolver o sistema linear pelo metodo de eliminacao deGauss com pivotacao parcial, usando 2 casas decimais:

2x1 − x2 + 2x3 = 2

3x1 + 2x2 − x3 = 1

2x2 + 4x3 = −1

.


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c©2013–2014 Loıc Cerf

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Eliminação de Gauss

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