Question

Resolver os sistemas por escapamentos e Cramer.

X+y +z = 6
2x-y+4z=5
3x+2y+z=13

Answer 1

Escalonamento
1 1 1 6
2 -1 4 5
3 2 1 13

1 1 1 6     L1 = (1*L1)
0 -3 2 -7  L2 = (-2*L1) + (1*L2)
0 -1 -2 -5 L3 = (-3*L1) + (1*L3)

1 1 1 6     L1 = (1*L1)
0 1 10 13 L2 = (1*L2) + (-4*L3)
0 -1 -2 -5  L3 = (1*L3)

1 1 1 6      L1 = (1*L1)
0 1 10 13  L2 = (1*L2)
0 0 8 8      L3 = (1*L2) + (1*L3)

1 1 1 6      L1 = (1*L1)
0 1 10 13  L2 = (1*L2)
0 0 1 1      L3 = (0,125*L3)

1 1 0 5      L1 = (1*L1) + (-1*L3)
0 1 0 3      L2 = (1*L2) + (-10*L3)
0 0 1 1      L3 = (1*L3)

1 0 0 2      L1 = (1*L1) + (-1*L2)
0 1 0 3      L2 = (1*L2)
0 0 1 1      L3 = (1*L3)


Cramer

Ma= 1 1 1 6
        2 -1 4 5
        3 2 1 13

M= 1 1 1 1 1
      2 -1 4 2 -1
      3 2 1 3 2
(1*-1*1+1*4*3+1*2*2)-(1*-1*3+1*4*2+1*2*1)
(-1+12+4)-(-3+8+2)
M=8

Mx= 6 1 1 6 1
        5 -1 4 5 -1
        13 2 1 13 2
Mx= (6*-1*1+1*4*13+1*5*2)-(1*-1*13+6*4*2+1*5*1) Mx= (-6+52+10)-(-13+48+5)
Mx= 16

My= 1 6 1 1 6
        2 5 4 2 5
        3 13 1 3 13
My= (1*5*1+6*4*3+1*2*13)-(1*5*3+1*4*13+6*2*1) My= (5+72+26)-(15+52+12)
My= 24

Mz= 1 1 6 1 1
        2 -1 5 2 -1
        3 2 13 3 2
Mz= (1*-1*13+1*5*3+6*2*2)-(6*-1*3+1*5*2+1*2*13) Mz= (-13+15+24)-(-18+10+26)
Mz= 8

x = Mx/M = 2
y = My/M = 3
z = Mz/M = 1