aol4 - calculo numerico

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Pergunta 1 -- /1
Calcule o determinante da matriz escalonada do sistema, utilizando o método de Eliminação de Gauss. Sendo o sistema: 
img_questao01NAD_calculo numerico_unidade03.PNG
22/5 
4/5 
Resposta correta-29 
-29/5 
10/10
ota final
Enviado: 25/09/20 10:03 (BRT)
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22 
Pergunta 2 -- /1
Dado o sistema linear:
open curly brackets table attributes columnalign left end attributes row cell table row cell 3 x subscript 1 plus end cell cell x subscript 2 minus end cell cell x 
subscript 3 equals end cell 2 row cell x subscript 1 plus end cell cell 5 x subscript 2 plus end cell cell 3 x subscript 3 equals end cell 15 end table end cell row cell 
table row cell 2 x subscript 1 plus end cell cell x subscript 2 plus end cell cell 4 x subscript 3 equals end cell 9 end table end cell end table close
Faça três iterações usando o Método de Gauss-Jacobi ( até Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. ). Para isso use 
como valores iniciais 
x to the power of 0 space equals space open square brackets table row cell table row cell 1 comma 000 end cell cell 1 comma 000 end cell end table end cell 
cell 1 comma 000 end cell end table space close square brackets to the power of tau
(realize os cálculos com três casas decimais).
Resposta correta
table row cell x cubed equals end cell cell open square brackets table row cell table row cell 0 comma 467 end cell cell 
space 2 comma 093 end cell end table end cell cell 1 comma 542 end cell end table space close square brackets to the power 
of tau end cell cell d cubed space equals space 0 comma 175 end cell end table
table row cell x cubed equals end cell cell open square brackets table row cell table row cell 0 comma 013 end cell cell 1 comma 876 end cell 
end table end cell cell 3 comma 451 end cell end table close square brackets to the power of tau end cell cell d cubed space equals space 0 
comma 087 end cell end table
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table row cell x cubed equals end cell cell open square brackets table row cell table row cell 0 comma 333 end cell cell space space 2 comma 
196 end cell end table end cell cell space 1 comma 631 end cell end table space close square brackets end cell cell d cubed space equals space 
0 comma 552 end cell end table
table row cell x cubed equals end cell cell open square brackets table row cell table row cell 2 comma 045 end cell cell 1 comma 003 end cell 
end table end cell cell 0 comma 610 end cell end table close square brackets to the power of tau end cell cell d cubed space equals space 0 
comma 017 end cell end table
table row cell x cubed equals end cell cell open square brackets table row cell table row cell 4 comma 012 end cell cell 0 comma 562 end cell 
end table end cell cell 2 comma 415 end cell end table close square brackets to the power of tau end cell cell d cubed space equals space 0 
comma 156 end cell end table
Pergunta 3 -- /1
Determine a solução do sistema linear:
open curly brackets table attributes columnalign left end attributes row cell table row cell x subscript 1 end cell cell plus space 2 x subscript 2 end cell cell 
minus space x subscript 3 end cell cell plus space x subscript 4 end cell cell equals 8 end cell row space cell 3 x subscript 2 end cell cell plus 2 x subscript 3 end 
cell cell plus space 4 x subscript 4 end cell cell equals 3 end cell row space space cell x subscript 3 end cell cell plus 3 x subscript 4 end cell cell equals 1 end 
cell end table end cell row cell table row space cell space space space space space space space space space end cell cell space space space space space 
space space space space space space space space space space space space space end cell cell 5 x subscript 4 end cell end table equals space 5 end cell end 
table close
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Resposta corretax subscript 1 space equals space 3 ; space x subscript 2 space equals space 1 ; space x subscript 3 space equals space 
minus 2 ; space x subscript 4 space equals space 1.
x subscript 1 space equals space minus 1 ; space x subscript 2 space equals space 0 ; space x subscript 3 space equals space 1 ; space x 
subscript 4 space equals space 2.
x subscript 1 space equals space 0 ; space x subscript 2 space equals space 2 ; space x subscript 3 space equals space 1 ; space x subscript 
4 space equals space 3.
x subscript 1 space equals space 1 ; space x subscript 2 space equals space 2 ; space x subscript 3 space equals space minus 4 ; space x 
subscript 4 space equals space 1.
x subscript 1 space equals space 5 ; space x subscript 2 space equals space minus 1 ; space x subscript 3 space equals space minus 4 ; 
space x subscript 4 space equals space 1.
Pergunta 4 -- /1
Aplique o Método de Eliminação Gaussiana para encontrar a solução do sistema seguinte.
open curly brackets table attributes columnalign left end attributes row cell table row cell 2 x subscript 1 end cell cell plus space 4 x subscript 2 end cell cell 
plus x subscript 3 end cell cell equals space 7 end cell row cell minus x subscript 1 end cell cell plus 3 x subscript 2 end cell cell plus 2 x subscript 3 end cell cell 
equals space 4 space end cell end table end cell row cell table row space cell 5 x subscript 1 end cell cell space space space space space space space space 
end cell cell table row cell plus 2 x subscript 3 end cell cell equals space 7 space end cell end table end cell end table end cell end table close
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Resposta corretatable row cell x space equals end cell cell open square brackets table row cell table row 1 1 end table end cell 1 end table 
close square brackets to the power of T to the power of blank end exponent end cell space space end table
table row cell x space equals end cell cell open square brackets table row cell table row cell minus 1 end cell 3 end table end cell 5 end table 
close square brackets to the power of T end cell space space end table
table row cell x space equals end cell cell open square brackets table row cell table row cell minus 1 end cell 2 end table end cell 1 end table 
close square brackets to the power of T end cell space space end table
table row cell x space equals end cell cell open square brackets table row cell table row 2 1 end table end cell 3 end table close square 
brackets to the power of T end cell space space end table
table row cell x space equals end cell cell open square brackets table row cell table row 0 cell minus 1 end cell end table end cell 3 end table 
close square brackets to the power of T end cell space space end table
Pergunta 5 -- /1
Determine a solução do sistema, utilizando o método de Eliminação de Gauss. Sendo o sistema: 
img_questao02NAD_calculo numerico_unidade03.PNG
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x1= 2, x2= 0, x3=0, x4= 1 
x1= 0, x2= -1, x3=0, x4= 1 
x1= - 2, x2= 1, x3=1, x4= 1
Resposta corretax1= - 2, x2= -1, x3=0, x4= 1 
x1= 2, x2= -1, x3=0, x4= -1 
Pergunta 6 -- /1
Aplique o Método de Eliminação Gaussiana para encontrar a soluçãodo sistema seguinte.
open curly brackets table attributes columnalign left end attributes row cell table row cell 4 x subscript 1 end cell cell plus space x subscript 2 end cell cell plus 
space 3 x subscript 3 end cell cell equals space 9 end cell row cell 5 x subscript 1 end cell cell minus space x subscript 2 end cell cell plus space 2 x subscript 3 
end cell cell equals space 9 end cell end table end cell row cell table row cell 7 x subscript 1 end cell cell plus space 2 x subscript 2 end cell cell plus 3 x 
subscript 3 end cell cell equals space 16 end cell end table end cell end table close
Resposta corretax space equals space open square brackets table row 2 1 0 end table close square brackets to the power of tau
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x space equals space open square brackets table row 0 1 3 end table close square brackets to the power of tau
x space equals space open square brackets table row 5 cell minus 2 end cell 1 end table close square brackets to the power of tau
x space equals space open square brackets table row cell minus 4 end cell cell minus 1 end cell 2 end table close square brackets to the 
power of tau
x space equals space open square brackets table row cell minus 1 end cell 2 1 end table close square brackets to the power of tau
Pergunta 7 -- /1
Dado o sistema seguinte, encontre a matriz triangular superior e a solução.
open curly brackets table attributes columnalign left end attributes row cell 2 x subscript 1 space minus space 3 x subscript 2 space equals space minus 11 
end cell row cell 4 x subscript 1 space plus space 5 x subscript 2 space equals space 11 end cell end table close
open square brackets table row 2 cell minus 3 end cell row 0 6 end table left enclose table row 11 row 3 end table end enclose close square 
brackets space e space x subscript 1 space equals space minus 3 ; space x subscript 2 space equals space 2.
open square brackets table row 2 cell minus 3 end cell row 0 5 end table left enclose table row 10 row 23 end table end enclose close square 
brackets space e space x subscript 1 space equals space 0 ; space x subscript 2 space equals space 4.
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open square brackets table row 1 cell minus 3 end cell row 0 1 end table left enclose table row cell minus 11 end cell row 8 end table end 
enclose close square brackets space e space x subscript 1 space equals space 2 ; space x subscript 2 space equals space 8.
Resposta correta
open square brackets table row 2 cell minus 3 end cell row 0 11 end table left enclose table row cell minus 11 end cell row 
33 end table end enclose close square brackets space e space x subscript 1 space equals space minus 1 ; space x subscript 
2 space equals 3.
open square brackets table row 2 3 row 0 7 end table left enclose table row 4 row 15 end table end enclose close square brackets space e 
space x subscript 1 space equals space minus 2 ; space x subscript 2 space equals space 5.
Pergunta 8 -- /1
O que se pode afirmar, analisando o sistema seguinte?
open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell 3 x subscript 1 end cell cell plus space 6 x subscript 2 space 
equals space 12 end cell row cell 12 x subscript 1 end cell cell plus space 24 x subscript 2 space end subscript equals space 48 end cell end table close
Resposta correta
O sistema apresenta infinitas soluções, pois o determinante da matriz dos coeficientes é nulo.
Assim, podemos encontrar uma das soluções dada por 
Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.
O sistema é incompatível. Assim, não apresenta solução.
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O sistema apresenta solução única, pois o determinante da matriz diferente de zero. 
Assim, a solução é dada por Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.
O sistema apresenta duas soluções possíveis, pois o determinante é igual a dois.
Assim, as soluções serão dadas por Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.
O sistema apresenta um número finito de soluções, pois o determinante da matriz dos coeficientes é menor que zero. Assim, obrigatoriamente 
todas as soluções encontradas terão valores negativos.
Pergunta 9 -- /1
Dado o sistema linear:
open curly brackets table attributes columnalign left end attributes row cell table row cell minus 4 comma 200 x subscript 1 plus end cell cell 1 comma 500 x 
subscript 2 space plus end cell cell 2 comma 200 x subscript 3 space equals end cell cell 4 comma 200 end cell row cell 3 comma 500 x subscript 1 space space 
space end subscript minus end cell cell 5 comma 500 x subscript 2 space plus end cell cell 0 comma 500 x subscript 3 space end subscript equals end cell cell 
minus 8 comma 500 end cell end table end cell row cell table row cell space space 0 comma 250 x subscript 1 space plus end cell cell 1 comma 200 x subscript 
2 space plus end cell cell 2 comma 500 x subscript 3 space equals end cell cell space 6 comma 257 end cell end table end cell end table close
Faça três iterações usando o Método de Gauss-Jacobi Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. Para isso use como 
valores iniciais 
table row cell x to the power of 0 space end exponent equals end cell cell open square brackets table row cell table row cell 1 comma 000 end cell cell space 
1 comma 000 end cell end table end cell cell 1 comma 000 end cell end table space close square brackets to the power of tau end cell end table
(realize os cálculos com três casas decimais).
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Resposta corretax cubed equals open square brackets space 0 comma 333 space space space 2 comma 196 space space space 1 comma 
631 space close square brackets space d cubed equals 0 comma 552
table row cell x cubed equals end cell cell open square brackets table row cell minus 1 comma 105 end cell cell 0 comma 412 end cell cell 3 
comma 514 end cell end table close square brackets to the power of tau end cell cell d cubed equals end cell cell 0 comma 508 end cell end 
table
table row cell x cubed equals end cell cell open square brackets table row cell 2 comma 411 end cell cell minus 0 comma 597 end cell cell 4 
comma 528 end cell end table close square brackets to the power of tau end cell cell d cubed equals end cell cell 0 comma 603 end cell end 
table
table row cell x cubed equals end cell cell open square brackets table row cell 0 comma 882 end cell cell 5 comma 021 end cell cell minus 1 
comma 423 end cell end table close square brackets to the power of tau end cell cell d cubed equals end cell cell 0 comma 541 end cell end 
table
table row cell x cubed end cell cell open square brackets table row cell 0 comma 7412 end cell cell minus 1 comma 319 end cell cell 2 comma 
402 end cell end table close square brackets to the power of tau space e end cell cell d cubed equals end cell cell 0 comma 127 end cell end 
table
Pergunta 10 -- /1
Usando o critério de convergência das linhas, indique quais valores de “ k“ o Método de Gauss-Jacobi vai gerar uma sequencia convergente para a solução do 
sistema.
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open curly brackets table attributes columnalign left end attributesrow cell table row cell kappa x subscript 1 plus end cell cell 3 x subscript 2 plus end cell cell 
x subscript 3 end cell cell equals space 1 end cell row cell kappa x subscript 1 plus end cell cell 6 x subscript 2 plus end cell cell x subscript 3 end cell cell equals 
space 2 end cell end table end cell row cell table row cell minus 2 x subscript 1 plus end cell cell 5 x subscript 2 plus space 8 x subscript 2 plus end cell cell 8 x 
subscript 3 end cell cell equals 3 end cell end table end cell end table close
table row kappa cell less than 6 end cell end table
Resposta corretatable row 4 cell less than kappa end cell cell less than 5 end cell end table
table row cell kappa space equals space 0 space o u end cell cell kappa greater than 2 end cell end table
table row kappa less than cell 1 space o u end cell kappa greater than 5 end table
kappa space equals space 7