Livro Algebra Linear Boldrini Pdf File

Livro Algebra Linear Boldrini Pdf File Average ratng: 3,9/5 2189 votes

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´Algebra Moderna Diego Oliveira - Vit´oria da Conquista/BA Livro: ´Algebra Linear - Editora Harbra (Boldrini/Costa/Figueiredo/Wetzler) diegoalvez @ pop.com.br Compilado dia Solucion´ario da 3 a edi¸c˜ao do livro de ´Algebra Linear dosautores: Jos´e Luiz Boldrini, Sueli I.Rodrigues Costa, Vera L´uciaFigueiredo e Henry G. Wetzler.Para quem desejar; uma c´opia do livro pode ser baixada em lin I mat2012 2/Algebra%20Linear%20Boldrini.pdf.A expectativa ´e que seja respondido um cap´ıtulo do livro pormˆes.

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Mas, infelizmente resolver e digitar (principalmente digitar),os exerc´ıcios desse livro leva um bom tempo. Assim, pode haveratrasos na postagem.

De todo modo, n˜ao deixe de acompanhar odocumento no link abaixo, para obter futuras atualiza¸c˜oes. EXERC´ICIO PAGINA 111. SejamA =  1 2 32 1 − 1 , B =  − 2 0 13 0 1 , C =  − 124  e D = [2, − 1]Encontre:a) A + Bb) A Cc) B Cd) C De) D Af) D Bg) − Ah) − D1 ´Algebra Moderna Diego Oliveira - Vit´oria da Conquista/BA Solu¸c˜ao de a: A + B  1 2 32 1 − 1  +  − 2 0 13 0 1  =  − 1 2 45 1 0  Solu¸c˜ao de b: A B  1 2 32 1 − 1   − 124  =  15 − 4  Solu¸c˜ao de g: − 1 A − 1  1 2 32 1 − 1  =  − 1 − 2 − 3 − 2 − 1 1  2. Seja A =  2 x 2 2 x − 1 0  Se A’ = A, ent˜ao x = Solu¸c˜ao: Se A’ = A ent˜ao:  2 2 x − 1 x 2 0  =  2 x 2 2 x − 1 0  Que resulta nas seguintes igualdades:2 = 2 e 2 x − 1 = x 2 Desta ultima igualdade tira-se que x = 1.3. Se A ´e uma matriz sim´etrica, ent˜ao A – A’. Solu¸c˜ao: Se A ´e sim´etrica ent˜ao A = A’ e portanto A – A’ = A – A = 0.

Assim, o resultado destaopera¸c˜ao seria uma matriz nula.2 ´Algebra Moderna Diego Oliveira - Vit´oria da Conquista/BA4. Se A ´e uma matriz triangular superior, ent˜ao A’ ´e. Solu¸c˜ao: Uma matriz triangular superior quando transposta passa a ser uma matriz triangular infe-rior.5. Se A ´e uma matriz diagonal, ent˜ao A’.

Solu¸c˜ao: Toda matriz diagonal ´e sim´etrica de modo que se A ´e uma matriz diagonal ent˜ao A’ = A.6. Classifique em verdadeiro ou falso:a) –A’ = –A’b) (A + B)’ = B’ + A’c) Se AB = 0, ent˜ao A = 0 ou B = 0d) k 1 A k 2 B = k 1 k 2 ABe) –A –B = –ABf) Se A e B s˜ao matrizes sim´etricas, ent˜ao AB = BAg) Se AB = 0, ent˜ao BA = 0h) Se ´e poss´ıvel efetuar o produto AA, ent˜ao A ´e matriz quadrada Solu¸c˜ao de a: Pela propriedade iv a proposi¸c˜ao ´e verdadeira. Solu¸c˜ao de b Pela propriedade iii a proposi¸c˜ao ´e verdadeira. Solu¸c˜ao de c: Falsa. Tomando A =  1 00 0  e B =  0 01 0  por exemplo, verifica-se que a proposi¸c˜aon˜ao ´e verdadeira. Solu¸c˜ao de d: Usando a associatividade3 ´Algebra Moderna Diego Oliveira - Vit´oria da Conquista/BA( k 1 k 2 ) AB = A ( k 1 k 2 ) B Usando a comutatividade A ( k 2 k 1 ) B = k 2 ( Ak 1 ) B = ( Ak 1 ) ( k 2 B ) = ( k 1 A )( k 2 B ). Solu¸c˜ao de e: Falsa.

Como contra exemplo tome A =  2 10 − 1  e B =  0 31 1  Solu¸c˜ao de f: Falsa. Como contra exemplo tome A =  2 00 1  e B =  2 11 0  Solu¸c˜ao de g: Falsa.

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