Resolva o sistema abaixo

[[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,4],[,,,2,,,,,6,,,,,3,-,2,|,,,,10],[-,4,-,12,,-,7,,,,,0,|,-,10],[,,,6,,,,18,,,,,11,,,,,0,|,,,,14]]]

Passo 1

Inicialmente vamos escalonar a matriz:

[[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,4],[,,,2,,,,,6,,,,,3,-,2,|,,,,10],[-,4,-,12,,-,7,,,,,0,|,-,10],[,,,6,,,,18,,,,,11,,,,0,|,,,,14]]] overset(l_2 = l_2 - l_1)(->) [[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,4],[,,,0,,,,,0,,,,,0,-,3,|,,,,,,6],[-,4,-,12,,-,7,,,,,0,|,-,10],[,,,6,,,,18,,,,,11,,,,0,|,,,,14]]] overset(l_3 = l_3 + 2 l_1)(->)

[[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,4],[,,,0,,,,,0,,,,,0,-,3,|,,,,,,6],[,,,0,,,,,0,,-,1,,,,,2,|,,-,2],[,,,6,,,,18,,,,,11,,,,0,|,,,,14]]] overset(l_4 = l_4 - 3 l_1)(->) [[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,4],[,,,0,,,,,0,,,,,0,-,3,|,,,,,,6],[,,,0,,,,,0,,-,1,,,,,2,|,,-,2],[,,,0,,,,,0,,,,,2,-,3,|,,,,,,2]]] overset(l_4 = l_4 + 2 l_3)(->)

[[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,,4],[,,,0,,,,,0,,,,,0,-,3,|,,,,,,,6],[,,,0,,,,,0,,-,1,,,,,2,|,,,-,2],[,,,0,,,,,0,,,,,0,,,,,1,,|,,,-,2]]] overset(l_2 harr l_3)(->) [[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,,4],[,,,0,,,,,0,,,-,1,,,,,2,|,,,,-,2],[,,,0,,,,,0,,,,,,0,-,3,|,,,,,,,6],[,,,0,,,,,0,,,,,,0,,,,,1,,|,,,-,2]]] overset(l_4 = l_4 + ((1)/(3)) l_3)(->)

[[[,,,,2,,,,,6,,,,,,3,,,,,1,|,,,,,,,4],[,,,0,,,,,0,,,-,1,,,,,2,|,,,,-,2],[,,,0,,,,,0,,,,,,0,-,3,|,,,,,,,6],[,,,0,,,,,0,,,,,,0,,,,,0,,|,,,,,,,0]]] overset(l_1 = ((1)/(2)) l_1)(->) [[[,,,,1,,,,,3,,,,,,(3)/(2),,,,,(1)/(2),|,,,,,,,2],[,,,0,,,,,0,,,-,1,,,,,2,|,,,,-,2],[,,,0,,,,,0,,,,,,0,-,3,|,,,,,,,6],[,,,0,,,,,0,,,,,,0,,,,,0,,|,,,,,,,0]]] overset(l_2 = - l_2)(->)

[[[,,,,1,,,,,3,,,,,,(3)/(2),,,,,(1)/(2),|,,,,,,,2],[,,,0,,,,,0,,,,,,1,-,2,|,,,,,,,2],[,,,0,,,,,0,,,,,,0,-,3,|,,,,,,,6],[,,,0,,,,,0,,,,,,0,,,,,0,,|,,,,,,,0]]] overset(l_1 = l_1 - ((3)/(2)) l_2)(->) [[[,,,,1,,,,,3,,,,,,0,,,,,(7)/(2),|,,,,,-,1],[,,,0,,,,,0,,,,,,1,,-,2,|,,,,,,,,2],[,,,0,,,,,0,,,,,,0,-,3,|,,,,,,,6],[,,,0,,,,,0,,,,,,0,,,,,0,,|,,,,,,,0]]] overset(l_3 = (- (1)/(3)) l_3)(->)

[[[,,,,1,,,,,3,,,,,,0,,,,,(7)/(2),|,,,,,-,1],[,,,0,,,,,0,,,,,,1,-,2,|,,,,,,,,2],[,,,0,,,,,0,,,,,,0,,,,1,|,,,,-,2],[,,,0,,,,,0,,,,,,0,,,,,0,,|,,,,,,,0]]] overset(l_1 = l_1 - ((7)/(2)) l_3)(->) [[[,,,,1,,,,,3,,,,,,0,,,,,0,|,,,,,,,6],[,,,0,,,,,0,,,,,,1,-,2,|,,,,,,,2],[,,,0,,,,,0,,,,,,0,,,,1,|,,,,-,2],[,,,0,,,,,0,,,,,,0,,,,,0,,|,,,,,,,0]]] overset(l_2 = l_2 + 2 l_3)(->)

[[[,,,,1,,,,,3,,,,,,0,,,,,0,|,,,,,,,6],[,,,0,,,,,0,,,,,,1,,,,,0,|,,,,-,2],[,,,0,,,,,0,,,,,,0,,,,1,|,,,,-,2],[,,,0,,,,,0,,,,,,0,,,,,0,,|,,,,,,,0]]]

Passo 2

Sabemos que o sistema relativo à matriz do enunciado é equivalente ao sistema relativo à matriz escalonada, então vamos resolver esse último sistema.

Sejam x_1 , x_2 , x_3 e x_4 as variáveis do sistema. Como os pivôs da matriz aparecem na primeira, terceira e quarta coluna segue que x_1 , x_3 e x_4 são variáveis dependentes, enquanto x_2 é uma variável livre. Definimos x_2 = t . Agora, vamos descobrir o valor das variáveis dependentes. Pela terceira equação do sistema temos que x_4 = - 2 , pela segunda temos que x_3 = - 2 e pela primeira temos x_1 + 3 x_2 = 6 => x_1 + 3 t = 6 => x_1 = 6 - 3 t . Logo, a solução do sistema é (6 - 3 t , t , - 2 , - 2) = (6 , 0 , - 2 , - 2) + t (- 3 , 1 , 0 , 0) = (6 , 0 , - 2 , - 2) + < (- 3,1 , 0 , 0) > .

Resposta

(6 , 0 , - 2 , - 2) + < (- 3,1 , 0 , 0) >