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Enunciado

Álgebra Linear e Geometria Analítica - Anton - Álgebra Linear- Ed: 10º - Capítulo 1.6 - Ex. 3

[[x_1,+,3,x_2,+,x_3,=,4],[2,x_1,+,2,x_2,+,x_3,=,-,1],[2,x_1,+,3,x_2,+,x_3,=,3]]

Passo 1

O enunciado fala pra gente resolver esse sistema invertendo a matriz de coeficientes e usando o Teorema 1.6.2 que diz que “Se A dor uma matriz invertível n xx n , então para cada matriz b de tamanho n xx 1 , o sistema de equações A x = b tem exatamente uma solução, a saber, x = A^(- 1) b ”. Então bora aplicar isso ai no nossos sistema.

Passo 2

Temos que

[[x_1,+,3,x_2,+,x_3,=,4],[2,x_1,+,2,x_2,+,x_3,=,-,1],[2,x_1,+,3,x_2,+,x_3,=,3]] => [[[1,3,1],[2,2,1],[2,3,1]]] [[[x_1],[x_2],[x_3]]] = [[[4],[-,1],[3]]] =>

=> underset(I)(underset(⏝)([[[1,3,1],[2,2,1],[2,3,1]]]^(- 1) [[[1,3,1],[2,2,1],[2,3,1]]])) [[[x_1],[x_2],[x_3]]] = [[[1,3,1],[2,2,1],[2,3,1]]]^(- 1) [[[4],[-,1],[3]]] =>

=> [[[x_1],[x_2],[x_3]]] = [[[1,3,1],[2,2,1],[2,3,1]]]^(- 1) [[[4],[-,1],[3]]]

Agora nossa matriz é de ordem maior que 2, então bora calcular a inversa dela através do algoritmo.

Passo 3

Nosso algoritmo diz pra operarmos [ A | I ] até chegarmos em [ I | X ] , onde X = A^(- 1) . Então bora lá

A = [[[1,3,1],[2,2,1],[2,3,1]]] => [A,I] = [[[[[1,3,1],[2,2,1],[2,3,1]]]],[[[[1,0,0],[0,1,0],[0,0,1]]]]] =>

=> [[],[L_2,=,L_2,-,2,L_1],[L_3,=,L_3,-,2,L_1]] = [[[[[1,3,1],[0,-,4,-,1],[0,-,3,-,1]]]],[[[[1,0,0],[-,2,1,0],[-,2,0,1]]]]] =>

=> [[],[L_2,=,L_2,-,L_3],[]] = [[[[[1,3,1],[0,-,1,0],[0,-,3,-,1]]]],[[[[1,0,0],[0,1,-,1],[-,2,0,1]]]]] =>

=> [[],[],[L_3,=,L_3,-,3,L_2]] = [[[[[1,3,1],[0,-,1,0],[0,0,-,1]]]],[[[[1,0,0],[0,1,-,1],[-,2,-,3,4]]]]] =>

=> [[,L_1,=,L_1,+,3,L_2],[],[]] = [[[[[1,0,1],[0,-,1,0],[0,0,-,1]]]],[[[[1,3,-,3],[0,1,-,1],[-,2,-,3,4]]]]] =>

=> [[,L_1,=,L_1,+,L_3],[],[]] = [[[[[1,0,0],[0,-,1,0],[0,0,-,1]]]],[[[[-,1,0,1],[0,1,-,1],[-,2,-,3,4]]]]] =>

=> [[,],[,L_2,=,-,L_2,],[L_3,=,-,L_3,]] = [[[[[1,0,0],[0,1,0],[0,0,1]]]],[[[[-,1,0,1],[0,-,1,1],[2,3,-,4]]]]] =>

Logo

A^(- 1) = [[[[[-,1,0,1],[0,-,1,1],[2,3,-,4]]]]]

Passo 4

Agora que encontramos a inversa da nossa matriz podemos finalmente resolver a equação

[[[x_1],[x_2],[x_3]]] = [[[1,3,1],[2,2,1],[2,3,1]]]^(- 1) [[[4],[-,1],[3]]] =>

=> [[[x_1],[x_2],[x_3]]] = [[[[[-,1,0,1],[0,-,1,1],[2,3,-,4]]]]] [[[4],[-,1],[3]]] = [[x_1,=,-,4,-,0,+,3,=,-,1],[x_2,=,0,+,1,+,3,=,4],[x_3,=,8,-,3,-,12,=,-,7]]

Logo x_1 = - 1 , x_2 = 4 e x_3 = - 7

Resposta

x_1 = - 1 , x_2 = 4 e x_3 = - 7

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