Álgebra Linear
e
Geometria Analítica
4ª aula
Sistemas de equações lineares
e
inversão de matrizes
Seja A uma matriz invertível
então
existe uma matriz B tal que AB = I
A B1
B2
AB1  I1
B3  Bn   I
AB2  I 2  ABn  I n
1 
0 
 
AB1  0
 

0
0 
1 
 
AB2  0
 

0
0 
0 
 
AB3  1 
 

0
…
0 
0 
 
ABn  0
 

1 
1 
0 
 
AB1  0
 

0
0 
1 
 
AB2  0
 

0
0 
0 
 
AB3  1 
 

0
…
0 
0 
 
ABn  0
 

1 
A determinação da inversa
faz-se resolvendo n sistemas
de equações todos com
matriz A
1 
0 
 
AB1  0
 

0
0 
1 
 
AB2  0
 

0
0 
0 
 
AB3  1 
 

0
…
0 
0 
 
ABn  0
 

1 
Todos os sistemas são
possíveis e determinados.
Tem que ser car(A) = n
Se A é nn e car(A) = n
então
a forma condensada de A é In.
Se A é nn e car(A) = n
então
a forma condensada de A é In.
Podem-se resolver os n
sistemas simultaneamente



A




1
0
0

0
1
0

0

0

1



0
0
0

0

0
0


1

Para determinar a inversa é
preciso condensar esta matriz



A




1
0
0

0
1
0

0

0

1



0
0
0

1

0
0


0

0
0

0
1
0

0
0

1



0

0
0

1
0

0
0


1




A1 




1 1 0 


A  1  1  1
1 0  1
1 1 0 


A  1  1  1
1 0  1
1 1 0 1 0 0


1  1  1 0 1 0
1 0  1 0 0 1
1 1 0 


A  1  1  1
1 0  1
1 1 0 1 0 0


1

1

1
0
1
0


1 0  1 0 0 1
1
1 0 1 0 0


0  2  1  1 1 0 
 1
0  1 0 0 1
1 1 0 


A  1  1  1
1 0  1
1 1 0 1 0 0


1

1

1
0
1
0


1 0  1 0 0 1
1
1 0 1 0 0


0  2  1  1 1 0 
0  1  1  1 0 1
1
1 0 1 0 0


0  2  1  1 1 0 
 1
0  1 0 0 1
1 1 0 


A  1  1  1
1 0  1
1 1 0 1 0 0


1

1

1
0
1
0


1 0  1 0 0 1
1
1 0 1 0 0


0  2  1  1 1 0 
 1
0  1 0 0 1
1
1 0 1 0 0


0  2  1  1 1 0 
0  1  1  1 0 1
1
1 0 1 0 0


0

1

1

1
0
1


0  2  1  1 1 0
1
1 0 1 0 0


0  1  1  1 0 1
0  2  1  1 1 0
1
1 0 1 0 0


1 1 1 0  1
0
0  2  1  1 1 0
1
1 0 1 0 0


0  1  1  1 0 1
0  2  1  1 1 0
1 1 0 1 0
0


0 1 1 1 0  1
0 0 1 1 1  2
1
1 0 1 0 0


1 1 1 0  1
0
0  2  1  1 1 0
1
1 0 1 0 0


0  1  1  1 0 1
0  2  1  1 1 0
1 1 0 1 0
0


0 1 1 1 0  1
0 0 1 1 1  2
1
1 0 1 0 0


1 1 1 0  1
0
0  2  1  1 1 0
1 1 0 1 0
0


1
0 1 0 0  1
0 0 1 1 1  2
1
1 0 1 0 0


0  1  1  1 0 1
0  2  1  1 1 0
1 1 0 1 0
0


0 1 1 1 0  1
0 0 1 1 1  2
 1 0 0 1 1  1


1
0 1 0 0  1
0 0 1 1 1  2
1
1 0 1 0 0


1 1 1 0  1
0
0  2  1  1 1 0
1 1 0 1 0
0


1
0 1 0 0  1
0 0 1 1 1  2
1
1 0 1 0 0


0  1  1  1 0 1
0  2  1  1 1 0
1
1 0 1 0 0


1 1 1 0  1
0
0  2  1  1 1 0
1 1 0 1 0
0


0 1 1 1 0  1
0 0 1 1 1  2
1 1 0 1 0
0


1
0 1 0 0  1
0 0 1 1 1  2
 1 0 0 1 1  1


1
0 1 0 0  1
0 0 1 1 1  2
 1 1  1
A1   0  1
1
 1 1  2
1 2 3 1 0 0 


 4 5 6 0 1 0
7 8 9 0 0 1
1 2
3 1 0 0


0  3  6  4 1 0
7 8
9 0 0 1
1
2
3 1 0 0 1 2
3 1
0 0

 

0  3  6  4 1 0   0  3  6  4 1 0 
0  6  12  7 0 1 0 0
0 1  2 1
A matriz não é invertível
2
0

0

0
0
6
0
0
0
0
8
0
0
0
0

4
2
0

0

0
0
6
0
0
0
0
8
0
0

0
0

4
0
0 
1 / 2 0
 0 1/ 6 0

0


 0
0 1/ 8 0 


0
0 1 / 4
 0
1 2 4 
0 1 2 


0 0 1 
1 2 4 1 0 0


0
1
2
0
1
0


0 0 1 0 0 1
1 2 0 1 0  4 1 0 0 1  2 0 

 

 0 1 0 0 1  2  0 1 0 0 1  2 
0 0 1 0 0 1  0 0 1 0 0
1 