eliminação de gauss
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Eliminacao de Gauss
Loıc Cerf
13 de fevereiro de 2014UFMG – ICEx – DCC
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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 − × 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
4 / 9Loıc Cerf Eliminacao de Gauss
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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
4 / 9Loıc Cerf Eliminacao de Gauss
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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
4 / 9Loıc Cerf Eliminacao de Gauss
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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
4 / 9Loıc Cerf Eliminacao de Gauss
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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
4 / 9Loıc Cerf Eliminacao de Gauss
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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
4 / 9Loıc Cerf Eliminacao de Gauss
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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
4 / 9Loıc Cerf Eliminacao de Gauss
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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
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
.
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
.
O vetor resıduo e: −202133
− − 1 1 − 2
5 2 12 5 4
3− 18
=
−202133
− − 20
2133
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
.
O vetor resıduo e: −202133
− − 1 1 − 2
5 2 12 5 4
3− 18
=
−202133
− − 20
21
33
A norma-2 do vetor resıduo e√
02 + 02 + 02 = 0.
6 / 9Loıc Cerf Eliminacao de Gauss
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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
− − 20
2133
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
.
O vetor resıduo e: −202133
− − 1 1 − 2
5 2 12 5 4
3− 18
=
000
A norma-2 do vetor resıduo e√
02 + 02 + 02 = 0.
6 / 9Loıc Cerf Eliminacao de Gauss
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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
=
000
A norma-2 do vetor resıduo e
√02 + 02 + 02 = 0.
6 / 9Loıc Cerf Eliminacao de Gauss
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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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License
c©2013–2014 Loıc Cerf
These slides are licensed under the Creative CommonsAttribution-ShareAlike 3.0 Unported License.
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