1  1 | 3 |  1
2 -1 | 0 | -7

L2 <- L2 -2 L1
ideia: zerar na 1ª coluna
abaixo da 1ª linha

1  1 | 3 |  1
0 -3 | -6| -9

L2 <- -1/3*L2

1  1 | 3 | 1
0  1 | 2 | 3

L1 <- L1 - L2

1  0 | 1     | -2
0  1 | 2     |  3

  I  | sol1  | sol2


A Y = I
a solução é Y = A^{-1}

  (A | I) ~ ... ~ (I | A^{-1})

 A^{-1}(A | I)
   = ( A^{-1} A | A^{-1} I  )
   = ( I | A^{-1})

================
A(n·m) B(m·k)

cliquei o botão a =(1,1)

0     1     1
0     1     1
0  +  0  =  0
1     1     0
1     0     1
0     0     0

clicar o botão b
1   1    0
1   1    0
0 + 1 =  1
0   0    0
1   1    0
0   0    0

1 1 0 1 0 0 A
1 1 1 0 1 0 B
0 1 1 0 0 1 C
1 0 0 1 1 0 D
0 1 0 1 1 1 E
0 0 1 0 1 1 F
~             L2 <- L2+L1
1 1 0 1 0 0 A
0 0 1 1 1 0 A+B
0 1 1 0 0 1 C
0 1 0 0 1 0 A+D
0 1 0 1 1 1 E
0 0 1 0 1 1 F
~
1 1 0 1 0 0 A
0 1 1 0 0 1 C
0 0 1 1 1 0 A+B
0 0 1 0 1 1 A+C+D
0 0 1 1 1 0 C+E
0 0 1 0 1 1 F
~ 
1 1 0 1 0 0 A
0 1 1 0 0 1 C
0 0 1 1 1 0 A+B     ****
0 0 0 1 0 1 B+C+D
0 0 0 0 0 0 A+B+C+E   
0 0 0 1 0 1 A+B+F   ****
(para o sistema ser possível,
A+B+C+E=0)
~
1 1 0 1 0 0 A
0 1 1 0 0 1 C
0 0 1 1 1 0 A+B    
0 0 0 1 0 1 B+C+D   
0 0 0 0 0 0 A+C+D+F 
0 0 0 0 0 0 A+B+C+E   
~
1 1 0 0 0 1 A+B+C+D 
0 1 1 0 0 1 C
0 0 1 0 1 1 A+C+D      ***
0 0 0 1 0 1 B+C+D    *** 
0 0 0 0 0 0 A+C+D+F 
0 0 0 0 0 0 A+B+C+E   
~
1 1 0 0 0 1 A+B+C+D 
0 1 0 0 1 0 A+D       ***
0 0 1 0 1 1 A+C+D   ***
0 0 0 1 0 1 B+C+D    
0 0 0 0 0 0 A+C+D+F 
0 0 0 0 0 0 A+B+C+E   
~
1 0 0 0 1 1 | B+C       ***
0 1 0 0 1 0 | A+D       ***
0 0 1 0 1 1 | A+C+D   
0 0 0 1 0 1 | B+C+D    
0 0 0 0 0 0 | A+C+D+F 
0 0 0 0 0 0 | A+B+C+E   

a = B + C + e + f

e=1
f=0
estado inicial:
(A,B,C,D,E,F)=(1,1,1,1,1,1)

a = 1 + 0 + 1 + 1     = 1
b = 1 + 1 + 1         = 1
c = 1 + 0 + 1 + 1 + 1 = 0
d = 0 + 1 + 1 + 1     = 1
e                     = 1
f                     = 0

apertaremos os botões a,b,d,e