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Página inicial / Meus cursos / EAD222004-75574 / Conteúdo / Exercício de Fixação - Tema 16 Iniciado em sexta, 18 Nov 2022, 18:57 Estado Finalizada Concluída em sexta, 18 Nov 2022, 18:58 Tempo empregado 1 minuto 36 segundos Avaliar 5,00 de um máximo de 5,00(100%) Questão 11 Correto Atingiu 1,00 de 1,00 Suponha a transformação , definida por F(x_1,x_2,x_3 )=(3x_1,〖2x〗_2+x_3) . Em R³, considere a base {e_1,e_2,e_3 }={(0,1,0),(1,0,0),(1,0,1)} . E em R², temos a base {h_1,h_2 }={(2,1),(1,0)} . Escolha uma opção: a. b. c. d. Sua resposta está correta. Vamos obter as transformações relativas à base {e_1,e_2,e_3 }={(0,1,0),(1,0,0),(1,0,1)} , e posteriormente verificar as combinações lineares que esta base realiza com a base {h_1,h_2 } . F(e_1 )=F(0,1,0)= F(x_1,x_2,x_3 )=(3x_1,〖2x〗_2+x_3) Substituindo os valores de (x_1,x_2,x_3) por (0,1,0) , gera-se o vetor em R²: F(e_1 )=(3*0,(2*1)+0)=(3,2) A partir da informação sobre a base {h_1,h_2 } , deve-se descrever e_1 e e_2 e e_3 como combinações lineares em função desta base. Assim, temos: F(e_1 )=F(1,0,0)→(3,2)=xh_1+yh_2→(3,2)=x(2,1)+y(1,0) Efetuando o sistema: Portanto, F(e_1 )=2h_1-1h_2 Agora, em e_2 F(e_2 )=F(1,0,0)=(3,0) (3,0)=x(2,1)+y(1,0) No sistema linear, temos: Assim, temos: F(e_2 )=0h_1+3h_2 Por fim, em e_3 : F(e_3 )=F(1,0,1)=(3,1) Exercício de Fixação - Tema 16: Revisão da tentativa https://ava.unicarioca.edu.br/ead/mod/quiz/review.php?attempt=1721601&cmid=265060 1 of 5 28/11/2022 14:07 https://ava.unicarioca.edu.br/ead/ https://ava.unicarioca.edu.br/ead/ https://ava.unicarioca.edu.br/ead/course/view.php?id=2701 https://ava.unicarioca.edu.br/ead/course/view.php?id=2701 https://ava.unicarioca.edu.br/ead/course/view.php?id=2701§ionid=58035#section-3 https://ava.unicarioca.edu.br/ead/course/view.php?id=2701§ionid=58035#section-3 https://ava.unicarioca.edu.br/ead/mod/quiz/view.php?id=265060 https://ava.unicarioca.edu.br/ead/mod/quiz/view.php?id=265060 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https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28e_3%20%29%3DF%281%2C0%2C1%29%3D%283%2C1%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28e_3%20%29%3DF%281%2C0%2C1%29%3D%283%2C1%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28e_3%20%29%3DF%281%2C0%2C1%29%3D%283%2C1%29%20 Questão 22 Correto Atingiu 1,00 de 1,00 Questão 33 Correto Atingiu 1,00 de 1,00 (3,1)=x(2,1)+y(1,0) No sistema linear, temos: x=1;y=1 Assim, temos: F(e_3 )=1h_1+1h_2 Portanto, a matriz representativa da transformação F é dada por: A resposta correta é: A fim de viabilizar os procedimentos de transformação linear, é importante recuperar alguns conceitos matemáticos, tais como o de imagem, domínio e contradomínio de uma função. As relações entre variáveis, específicas das funções que envolvem elementos de cálculo (como as funções logarítmicas, quadráticas e polinomiais, por exemplo) são conceitos que se aplicam também aos espaços vetoriais euclidianos. SANTANA, Ana Paula; QUEIRÓ, João Filipe. Introdução à Álgebra Linear. Lisboa: Gradiva, 2010. A partir do conteúdo proposto, analise as afirmativas que se seguem: I. ∀u_1,u_2∈U,F(u_1)=F(u_2),∀ u_1=u_2 (função bijetora). II. Se G = Im(z), ∀g∈G,há z∈Z,∴ , F(z) = g (função sobrejetora). III. Pela regra da função injetora, ∀k_n,k_m∈K,F(k_n)=F(k_m),∀ n≠m . IV. Toda função bijetora é sobrejetora, e qualquer função injetora é bijetora. Agora, assinale a opção que contenha a(s) afirmativa(s) correta(s): Escolha uma opção: a. Apenas II e III. b. Apenas I e IV. c. Apenas II. d. Apenas I, III e IV. Sua resposta está correta. A segunda afirmativa está correta, pois a regra da função sobrejetora nos mostra a correspondência entre cada elemento do domínio e da imagem de uma função, conceito que será usado também para as transformações lineares. Deste modo, se uma variável é uma imagem (contradomínio) da outra, como G é imagem de Z, para qualquer elemento g que pertença a este contradomínio existe um elemento z no domínio da função, de modo que F(z) = Z. A resposta correta é: Apenas II. Considere a transformação F: R³ \Rightarrow R² , definida por F(x_1,x_2,x_3 )=(x_2,x_1+x_3) . Em R³, considere a base canônica {e_1,e_2,e_3 }={(1,0,0),(0,1,0),(0,0,1)} . E em R², há a base {h_1,h_2 }={(2,1),(1,0)} . Nestas condições, qual a matriz representativa da transformação F? Escolha uma opção: a. Rio Comprido Av. Paulo de Frontin, 568 Rio Comprido, Rio de Janeiro, RJ Méier Rua Venceslau, 315 Méier, Rio de Janeiro, RJ Central de atendimento: (21) 2563-1919 © 2018 UniCarioca | Todos os direitos reservados. Exercício de Fixação - Tema 16: Revisão da tentativa https://ava.unicarioca.edu.br/ead/mod/quiz/review.php?attempt=1721601&cmid=265060 2 of 5 28/11/2022 14:07 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%283%2C1%29%3Dx%282%2C1%29%2By%281%2C0%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%283%2C1%29%3Dx%282%2C1%29%2By%281%2C0%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%283%2C1%29%3Dx%282%2C1%29%2By%281%2C0%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%283%2C1%29%3Dx%282%2C1%29%2By%281%2C0%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20x%3D1%3By%3D1%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20x%3D1%3By%3D1%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20x%3D1%3By%3D1%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20x%3D1%3By%3D1%20 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https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%7Bh_1%2Ch_2%20%7D%3D%7B%282%2C1%29%2C%281%2C0%29%7D%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%7Bh_1%2Ch_2%20%7D%3D%7B%282%2C1%29%2C%281%2C0%29%7D%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%7Bh_1%2Ch_2%20%7D%3D%7B%282%2C1%29%2C%281%2C0%29%7D%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%7Bh_1%2Ch_2%20%7D%3D%7B%282%2C1%29%2C%281%2C0%29%7D%20 https://ava.unicarioca.edu.br/ead/mod/quiz/review.php?attempt=1721601&cmid=265060# https://ava.unicarioca.edu.br/ead/mod/quiz/review.php?attempt=1721601&cmid=265060# Questão 44 Correto Atingiu 1,00 de 1,00 b. c. d. Sua resposta está correta. Devemos calcular primeiro as transformações relativas à base canônica, para depois verificarmos as combinações lineares que a base canônica exerce com a base de dimensão R². Portanto: F(e_1 )=F(1,0,0)= F(x_1,x_2,x_3 )=(x_2,x_1+x_3) Substituindo os valores de (x_1,x_2,x_3) por (1,0,0) , gera-se o vetor em R²: F(e_1 )=(0,1+0)=(0,1) A partir da informação sobre a base {h_1,h_2 } , deve-se descrever e_1 e e_2 e e_3 como combinações lineares em função desta base, assim, temos: F(e_1 )=F(1,0,0)=(0,1)=ah_1+bh_2 Você pode encontrar esta resposta efetuando um sistema linear: A resposta correta é: Quando a relação entre espaços vetoriais envolve mais de uma relação de correspondência entre domínio e imagem (ou contradomínio), pode-se afirmar a existência de uma composição de transformações lineares, onde uma transformação se expressa em função de outra. CALLIOLI, Carlos Alberto; DOMINGUES, Hygino Hugueros; COSTA, Roberto Celso Fabrício. Álgebra Linear e Aplicações. 6. ed, 19 reimpr. São Paulo: Atual, 2013. Com base no exposto, considere a composição . Sob quais condições este processo é uma transformação linear? Escolha uma opção: a. (H o J)(β^2 x)= (β(H o J)(x))^2. . b. (H o J)(β^2 x)=J(Hβ^2 x) c. (H o J)(β^2 x)≠β^2 (H o J)(x). Exercício de Fixação - Tema 16: Revisão da tentativa https://ava.unicarioca.edu.br/ead/mod/quiz/review.php?attempt=1721601&cmid=265060 3 of 5 28/11/2022 14:07 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28e_1%20%29%3DF%281%2C0%2C0%29%3D%20F%28x_1%2Cx_2%2Cx_3%20%29%3D%28x_2%2Cx_1%2Bx_3%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28e_1%20%29%3DF%281%2C0%2C0%29%3D%20F%28x_1%2Cx_2%2Cx_3%20%29%3D%28x_2%2Cx_1%2Bx_3%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28e_1%20%29%3DF%281%2C0%2C0%29%3D%20F%28x_1%2Cx_2%2Cx_3%20%29%3D%28x_2%2Cx_1%2Bx_3%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28e_1%20%29%3DF%281%2C0%2C0%29%3D%20F%28x_1%2Cx_2%2Cx_3%20%29%3D%28x_2%2Cx_1%2Bx_3%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28x_1%2Cx_2%2Cx_3%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28x_1%2Cx_2%2Cx_3%29%20 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https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%3DJ%28H%CE%B2%5E2%20x%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%3DJ%28H%CE%B2%5E2%20x%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%3DJ%28H%CE%B2%5E2%20x%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%3DJ%28H%CE%B2%5E2%20x%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%E2%89%A0%CE%B2%5E2%20%28H%20o%20J%29%28x%29.%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%E2%89%A0%CE%B2%5E2%20%28H%20o%20J%29%28x%29.%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%E2%89%A0%CE%B2%5E2%20%28H%20o%20J%29%28x%29.%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%E2%89%A0%CE%B2%5E2%20%28H%20o%20J%29%28x%29.%20 Questão 55 Correto Atingiu 1,00 de 1,00 c. (H o J)(β^2 x)≠β^2 (H o J)(x). d. (H o J)(β^2 x)=β^2 (H o J)(x). Sua resposta está correta. A resposta correta é: (H o J)(β^2 x)=β^2 (H o J)(x). Supondo a existência de um espaço vetorial R, organizado pelos vetores dispostos na base ( B={r_1,r_2,…,r_n} \), e a base C={s_1,s_2,…,s_n} , associada ao espaço S que forma a imagem de R, descreva a forma geral que demonstra a transformação F(r_j ) : Escolha uma opção: a. b. c. \phi d. Sua resposta está correta. Como o espaço vetorial S é a imagem do espaço R, formando assim o seu contradomínio, a transformação linear ocorre em função de R, para todos os n elementos de R, que formam uma combinação linear com os elementos de S. Portanto, temos: F(r_1 )=α_11 s_1+α_12 s_2+⋯+α_1m s_m F(r_2 )=α_21 s_1+α_22 s_2+⋯+α_2m s_m F(r_n )=α_n1 s_1+α_n2 s_2+⋯+α_nm s_m O processo F(r_n ) nos mostra que cada elemento de R é uma combinação linear de S, por meio da soma dos produtos entre coeficientes lineares (que podem ser dispostos na forma matricial, com n linhas e m colunas) e as coordenadas vetoriais dos Exercício de Fixação - Tema 16: Revisão da tentativa https://ava.unicarioca.edu.br/ead/mod/quiz/review.php?attempt=1721601&cmid=265060 4 of 5 28/11/2022 14:07 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%E2%89%A0%CE%B2%5E2%20%28H%20o%20J%29%28x%29.%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%E2%89%A0%CE%B2%5E2%20%28H%20o%20J%29%28x%29.%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20%28H%20o%20J%29%28%CE%B2%5E2%20x%29%E2%89%A0%CE%B2%5E2%20%28H%20o%20J%29%28x%29.%20 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https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_1%20%29%3D%CE%B1_11%20s_1%2B%CE%B1_12%20s_2%2B%E2%8B%AF%2B%CE%B1_1m%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_1%20%29%3D%CE%B1_11%20s_1%2B%CE%B1_12%20s_2%2B%E2%8B%AF%2B%CE%B1_1m%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_2%20%29%3D%CE%B1_21%20s_1%2B%CE%B1_22%20s_2%2B%E2%8B%AF%2B%CE%B1_2m%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_2%20%29%3D%CE%B1_21%20s_1%2B%CE%B1_22%20s_2%2B%E2%8B%AF%2B%CE%B1_2m%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_2%20%29%3D%CE%B1_21%20s_1%2B%CE%B1_22%20s_2%2B%E2%8B%AF%2B%CE%B1_2m%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_2%20%29%3D%CE%B1_21%20s_1%2B%CE%B1_22%20s_2%2B%E2%8B%AF%2B%CE%B1_2m%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%3D%CE%B1_n1%20s_1%2B%CE%B1_n2%20s_2%2B%E2%8B%AF%2B%CE%B1_nm%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%3D%CE%B1_n1%20s_1%2B%CE%B1_n2%20s_2%2B%E2%8B%AF%2B%CE%B1_nm%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%3D%CE%B1_n1%20s_1%2B%CE%B1_n2%20s_2%2B%E2%8B%AF%2B%CE%B1_nm%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%3D%CE%B1_n1%20s_1%2B%CE%B1_n2%20s_2%2B%E2%8B%AF%2B%CE%B1_nm%20s_m%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%20https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20F%28r_n%20%29%20 coeficientes lineares (que podem ser dispostos na forma matricial, com n linhas e m colunas) e as coordenadas vetoriais dos elementos de S(s_1,s_2,…,s_n) . Portanto, temos: A resposta correta é: ◄ Vídeo - Tema 16 Seguir para... Apresentação - Tema 17 ► Exercício de Fixação - Tema 16: Revisão da tentativa https://ava.unicarioca.edu.br/ead/mod/quiz/review.php?attempt=1721601&cmid=265060 5 of 5 28/11/2022 14:07 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20S%28s_1%2Cs_2%2C%E2%80%A6%2Cs_n%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20S%28s_1%2Cs_2%2C%E2%80%A6%2Cs_n%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20S%28s_1%2Cs_2%2C%E2%80%A6%2Cs_n%29%20 https://ava.unicarioca.edu.br/ead/filter/tex/displaytex.php?texexp=%20S%28s_1%2Cs_2%2C%E2%80%A6%2Cs_n%29%20 https://ava.unicarioca.edu.br/ead/mod/url/view.php?id=265059&forceview=1 https://ava.unicarioca.edu.br/ead/mod/url/view.php?id=265059&forceview=1 https://ava.unicarioca.edu.br/ead/mod/scorm/view.php?id=265062&forceview=1 https://ava.unicarioca.edu.br/ead/mod/scorm/view.php?id=265062&forceview=1
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