Seja A = [[[1,0,5],[1,1,1],[0,1,-,4]]]

b) Encontre a solução geral do sistema (A-2 I_3 ) X = 0

Passo 1

Inicialmente:

I_3 = [[[1,0,0],[0,1,0],[0,0,1]]]

2 I_3 = [[[2,0,0],[0,2,0],[0,0,2]]]

A-2 I_3 = [[[1,-,2,0,-,0,5,-,0],[1,-,0,1,-,2,1,-,0],[0,-,0,1,-,0,-,4,-,2]]]

A-2 I_3 = [[[-,1,0,5],[1,-,1,1],[0,1,-,6]]]

Passo 2

Montando um sistema em forma de matriz e resolvendo:

[[[-,1,0,5],[1,-,1,1],[0,1,-,6]],[[0],[0],[0]]]

[[[-,1,0,5],[1,-,1,1],[0,1,-,6]],[[0],[0],[0]]] , L_1 // ( - 1 ) -> L_1 :

[[[1,0,-,5],[1,-,1,1],[0,1,-,6]],[[0],[0],[0]]] , L_2 - 1 L_1 -> L_2 :

[[[1,0,-,5],[0,-,1,6],[0,1,-,6]],[[0],[0],[0]]] , L_2 // ( - 1 ) -> L_2 :

[[[1,0,-,5],[0,1,-,6],[0,1,-,6]],[[0],[0],[0]]] , L_3 - ( 1 ) L_2 -> L_3 :

[[[1,0,-,5],[0,1,-,6],[0,0,0]],[[0],[0],[0]]]

Passo 3

{[[x_1 - 5 x_3 = 0],[x_2 - 6 x_3 = 0],[]])

Resposta

X = ([[5,a],[6,a],[a]])

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