Resolva, usando a Regra de Cramer:
a) l3 x-y+z=1 2 x+3 z=-1 4 x+y-2 z=7 c) r-x-y+z=1 x+y+z=1 2 x+3 y+2 z=0
b) x-y+z & =-5 x+2 y+4 z & =4 3 x+y-2 z & =-3
apêndice A

Passo 1

Temos três sistemas lineares 3times 3 e vamos resolvê-los usando a Regra de Cramer. Isso envolve montar a matriz dos coeficientes A, calcular o determinante D=det A e depois, para cada incógnita, substituir a coluna correspondente pelos termos independentes, obtendo determinantes Dx,Dy,Dz. Cada incógnita vem de ,x=Dx/D, ,y=Dy/D e ,z=Dz/D.

Passo 2

Sistema (a):
3x - y + z = 1 2x +3z = -1 4x + y -2z = 7
Matriz dos coeficientes e determinante principal:
D = 3 & -1 & 1 2 & 0 & 3 4 & 1 & -2 =3,(0cdot(-2)-3cdot1);-;(-1),(2cdot(-2)-3cdot4);+;1,(2cdot1-0cdot4) =-9-16+2=-23
Determinantes substituindo colunas:
Dx= 1 & -1 & 1 -1& 0 & 3 7 & 1 & -2 =-3-19-1=-23, Dy= 3 & 1 & 1 2 & -1& 3 4 & 7 & -2 =-57+16+18=-23,
Dz= 3 & -1 & 1 2 & 0 & -1 4 & 1 & 7 =3+18+2=23
Logo
x=DxD=1, y=DyD=1, z=DzD=-1

Passo 3

Sistema (b):
x - y + z = -5 x +2y+4z = 4 3x+y -2z = -3
D = 1 & -1 & 1 1 & 2 & 4 3 & 1 & -2 =1cdot(2cdot(-2)-4cdot1)-(-1)cdot(1cdot(-2)-4cdot3)+1cdot(1cdot1-2cdot3) =-8-14-5=-27
Dx= -5 & -1 & 1 4 & 2 & 4 -3 & 1 & -2 =40+4+10=54, Dy= 1 & -5 & 1 1 & 4 & 4 3 & -3 & -2 =4-70-15=-81,
Dz= 1 & -1 & -5 1 & 2 & 4 3 & 1 & -3 =-10-15+25=0
Logo
x=54-27=-2, y=-81-27=3, z=0-27=0

Passo 4

Sistema (c):
-x - y + z = 1 x + y + z = 1 2x+3y+2z = 0
D = -1 & -1 & 1 1 & 1 & 1 2 & 3 & 2 =-1,(1cdot2-1cdot3)-(-1),(1cdot2-1cdot2)+1,(1cdot3-1cdot2) =1+0+1=2
Dx= 1 & -1 & 1 1 & 1 & 1 0 & 3 & 2 =-1+2+3=4, Dy= -1 & 1 & 1 1 & 1 & 1 2 & 0 & 2 =-2+0+2=0,
Dz= -1 & -1 & 1 1 & 1 & 1 2 & 3 & 0 =-3+2+5=4
Logo
x=42=2, y=02=0, z=42=2

Resposta

a) ,(x,y,z)=(1,1,-1)
b) ,(x,y,z)=(-2,3,0)
c) ,(x,y,z)=(2,0,2)

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