Resolver os seguintes sistemas de Cramer:

C) {[[x - y + z + t = 0],[x + y - z + t = 1],[- x + y + z - t = 0],[2 x - y - z + 3 t = 1]])

Passo 1

Antes de mais nada vamos reconhecer quem será a matriz A , X e B .

A = ([[1,-,1,1,1],[1,1,-,1,1],[-,1,1,1,-,1],[2,-,1,-,1,3]]) , X = ([[x],[y],[z],[t]]) e B = ([[0],[1],[0],[1]])

Assim para encontrar os valores de x e y usaremos

X = A^(- 1) B

Beleza? Então para isso precisamos encontrar a matriz A^(- 1)

Passo 2

Para determinar a inversa da matriz A

([[1,-,1,1,1],[1,1,-,1,1],[-,1,1,1,-,1],[2,-,1,-,1,3]],[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]])

vamos fazer algumas operações. Iremos numerá-las aqui e em seguida colocar como ficará a matriz após cada operação.

Operação 1:

L_1 - L_2 -> L_2

L_1 + L_3 -> L_3

L_4 - 2 L_1 -> L_4

([[1,-,1,1,1],[0,-,2,2,0],[0,0,2,0],[0,1,-,3,1]],[[1,0,0,0],[1,-,1,0,0],[1,0,1,0],[-,2,0,0,1]])

Operação 2:

2 L_4 + L_2 -> L_4

([[1,-,1,1,1],[0,-,2,2,0],[0,0,2,0],[0,0,-,6,2]],[[1,0,0,0],[1,-,1,0,0],[1,0,1,0],[-,3,-,1,0,2]])

Operação 3:

3 L_3 + L_4 -> L_4

([[1,-,1,1,1],[0,-,2,2,0],[0,0,2,0],[0,0,0,2]],[[1,0,0,0],[1,-,1,0,0],[1,0,1,0],[0,-,1,3,2]])

Operação 4:

L_2 - L_3 -> L_2

([[1,-,1,1,1],[0,-,2,0,0],[0,0,2,0],[0,0,0,2]],[[1,0,0,0],[0,-,1,-,1,0],[1,0,1,0],[0,-,1,3,2]])

Operação 5:

L_2 - 2 L_1 -> L_1

([[-,2,0,-,2,-,2],[0,-,2,0,0],[0,0,2,0],[0,0,0,2]],[[-,2,-,1,-,1,0],[0,-,1,-,1,0],[1,0,1,0],[0,-,1,3,2]])

Operação 6:

L_1 + L_3 -> L_1

([[-,2,0,0,-,2],[0,-,2,0,0],[0,0,2,0],[0,0,0,2]],[[-,1,-,1,-,2,0],[0,-,1,-,1,0],[1,0,1,0],[0,-,1,3,2]])

Operação 7:

L_1 + L_4 -> L_1

([[-,2,0,0,0],[0,-,2,0,0],[0,0,2,0],[0,0,0,2]],[[-,1,-,2,1,2],[0,-,1,-,1,0],[1,0,1,0],[0,-,1,3,2]])

Operação 8:

(- 1)/(2) L_1 -> L_1

(- 1)/(2) L_2 -> L_2

(1)/(2) L_3 -> L_3

(1)/(2) L_4 -> L_4

([[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]],[[1 // 2,1,- 1 // 2,-,1],[0,1 // 2,1 // 2,0],[1 // 2,0,1 // 2,0],[0,- 1 // 2,3 // 2,1]])

Assim encontramos

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

Passo 3

Para encontrar os valores de x e y usaremos

X = A^(- 1) B

Ficará assim

([[x],[y],[z],[t]]) = ([[1 // 2,1,- 1 // 2,-,1],[0,1 // 2,1 // 2,0],[1 // 2,0,1 // 2,0],[0,- 1 // 2,3 // 2,1]]) ([[0],[1],[0],[1]]) = ([[0],[1 // 2],[0],[1 // 2]])

Logo,

x = 0 , y = 1 // 2 , z = 0 e t = 1 // 2

Resposta

x = 0 , y = 1 // 2 , z = 0 e t = 1 // 2

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