Respostas Capítulo 4_EV_Boldrini_2ª parte_2021

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2 3 3 2
0 1 2 3 (1 ) (1 ) (1 ) (1)a a t a t a t a t b t c t d+ + + = − + − + − +
3 22a at b bt bt c ct d= − + − + + − +
2 3( ) ( 2 )a b c d b c t bt at= + + + + − − + −
0
1
2
3
2
a b c d a
b c a
b a
a a
+ + + =

− − =

=
− =
3
2
1 1 22 2
a a
b a
b c a c a a
= −

=
− − =  = − −
3 2
3 2 1 2 0 1 2 3(1 ) (1 2 ) ( 2 )(1 ) ( )(1)a t a t t a a t a a a a= − − + − + + − − − + + + +
0 3 2 1 2 0( 2 )d a b c a d a a a a a= − − − +  = − − − − − +
3 2
3 3 2 2 2 1 1 2 2 0 1 2 32 2 2a a t a a t a t a a t a a t a a a a= − + + − + − + − + + + + +
3 2
3 2 1 0a t a t a t a= + + +
2 3 3
2 3 2
2
2 3
1 0 0 1
1 2 0 (1 )
1 0 0 1
1 0 0 0 1
t t t t
t t t t
t t t t
t t t
+ + − = −
− + + = −
− + + = −
+ + + =
1 0 0 1
1 2 1 0
1 1 0 0
1 0 0 0
− 
 
− 
 −
 
 
1 5 1 1 2 4 1 7 0 0
4 2 1 5 5 7 5 1 0 0
x y z t
− − −         
+ + + =         
− − − −         
2 0
5 4 7 0
4 5 5 0
2 5 7 0
x y z t
x y z t
x y z t
x y z t
+ + + =

− + − − =

− − − − =
 + + + =
2 1 2
3 1 3
4 1 4
1 1 2 1 0
55 1 4 7 0
44 1 5 5 0
22 5 7 1 0
L L L
L L L
L L L
 
 
→ +− − − 
  → +− − − −
 
→ − + 
3 2
4 2
1 1 2 1 0
0 6 6 2 0
20 3 3 1 0
20 3 3 1 0
L L
L L
 
 
− 
  =−
 
=− 
1 1 2 1 0
1
0 1 1 0
3
0 0 0 0 0
0 0 0 0 0
 
 
 −
 
 
 
 
 
p=2 e n = 4 São LD
2 1 2
3 1 3
4 1 4
1 5 4 2
1 1 1 5
22 4 5 7
1 7 5 1
L L L
L L L
L L L
− − 
 
→ − +− 
  → − +− −
 
→ − +− − 
Utilizando o algoritmo linha (Lipshutz)
2 3
4 2
1 5 4 2
0 6 3 3
0 0 0 0 3
0 0 0 0
L L
L L
− − 
 
= 
 − =
 
 
1 5 0 2
,
4 2 1 1

 −   
=     
−    
Dim(W) = 2
1 2 3 4(2, 3,2,2)
2 1 0 2 1
3 1 0 2 0
2 0 1 1 0
2 0 1 1 0
a b c d
a b c d
− = + + +
−         
         
− −         = + + +
         
         
         
v v v v
2 1 2
3 4
1 0 2 1 2
1 0 2 0 3
0 1 1 0 2
0 1 1 0 2
L L L
L L
− 
 
→ +− −  
 
 
= 
1 0 2 1 2
0 0 0 1 1
0 1 1 0 2
0 0 0 0 0
− 
 
− 
 
 
 
2 3
1 0 2 1 2
0 0 0 1 1
0 1 1 0 2
0 0 0 0 0
L L
− 
 
−   
 
 
 
1 0 2 1 2
0 1 1 0 2
0 0 0 1 1
0 0 0 0 0
− 
 
 
 −
 
 
1
2
2 3
d
b c
a c
= −

= −
 = +
2 1 0 2 1
3 1 0 2 0
(2 3) (2 ) ( 1)
2 0 1 1 0
2 0 1 1 0
c c c
−         
         
− −         = + + − + + −
         
         
         
Logo,  1 2 3 4(2, 3,2,2) , , ,−  v v v v
2 4
1 1 0 0 1 1 0 0
0 0 1 1 1 0 0 0
2 2 1 1 2 2 1 1
1 0 0 0 0 0 1 1
L L
− −   
   
    
   − −
   
   
b) 
Utilizando o algoritmo do Espaço Linha
3 4
1 1 0 0 1 1 0 0
0 1 0 0 0 1 0 0
0 0 1 1 0 0 1 1
0 0 1 1 0 0 0 0
L L
− −   
   
   
   =
   
   
{(1, 1,0,0), (0,1,0,0), (0,0,1,1)}W = − ( ) 3Dim W =
c) Não. Porque a Dim(4) = 4 e a Dim(W) = 3, apesar que 
W  4
Utilizando o algoritmo do Espaço Linha
3 1
1 1 0
0 1 1
1 1 1 L L
 
 
− 
  → − 
1 1 0
~ 0 1 1
0 0 1
 
 
− 
 
 
1 2 33 [ , , ] 3p Dim=  =v v v
Ou
(1,1,0) (0, 1,1) (1,1,1) (0,0,0)a b c+ − + =
2 1 2
1 0 1 0
1 1 1 0
0 1 1 0 L L L
 
 
−  
  → − +  3 2 3
1 0 1 0
0 1 0 0
0 1 1 0 L L L
 
 
 
  → − + 
1 0 1 0
0 1 0 0
0 0 1 0
 
 
 
 
 
0 . .a b c L I= = = 
1 1 4 0
2 4 6 0
0 6 14 0
 −
 
−  
 − 
1 1 4 0
0 6 14 0
0 6 14 0
 −
 
−  
 − 
1 1 4 0
0 6 14 0
0 0 0 0
 −
 
− 
 
 
0a b c= = =
O Sistema tem ser homogêneo 
2 e 3p n SPI= = 2 3
2 3
2 3
6 14 0
6 14
7
3
x x
x x
x x
− =
=
=
1 2 3
1 2 3
1 3 3 3
4 0
4
7 5
4
3 3
x x x
x x x
x x x x
− + =
= −
= − = − 3 3 3 3
5 7
, , /
3 3
W x x x x
  
= −   
  
( ) 1 e a nulidade 3 2 1Dim W n p= − = − =
1 0 0
1 1 0
0 1 1
U
W
W
 
 
 
 
 
Utilizando o algoritmo do Espaço Linha
1 0 0
0 1 0
0 0 1
 
 
 
 
 
1 0 0
0 1 0
0 1 1
 
 
 
 
 
3
3
( ) 3
( )
p
Dim U W
Dim U W
=
+ =
+ =
3 U W= 
1 2 1 2
1 2 1 2
Note que , pois , visto que, por definição
 e , logo .
W W W W
W W W W
    
   
0
0 0 0
1 2 1 2
1 2 1 2
Suponha inicialmente que:
Se e 
 e 
W W W W
W W W W
    
    
u u u
v v v
1 2 1 2
1 2 1 2
Com isso, temos se , e , , e como e 
são subespaços vetorias, tem-se:
 e . Logo, 
W W W W
W W W W
 
+  +  +  
u v u v
u v u v u v
1 2 1 2
1 2 1 2
1 2
Suponha agora que:
Se e e 
como e são subespaços vetorias, temos então que: .
Logo, é um subespaço vetorial de .
W W W W
W W W W
W W V
   

     
 

u u u
u
1 2 1 1 2 1
1 2 2 1 2 2
Suponha inicialmente que:
Se e 
 e 
W W
W W
  + 
  + 
u u u u
v v v v
1 2
1 1 2 2 2
1 1 2 2 1 2 1 2 1 2
Suponha que ,então :
 e 
Assim: 
 ( ) ( ) ( ) ( ) + .
W W W
W W
W W W
= +
+  + 
+ + + = + + +  =
u v u v
u v u v u u v v
1 1 1 2
1 1
1 1 1 1 1 2
Suponha agora que:
Se e e 
Como ( ) , então:
( )
 é um subespaço vetorial.
W W
W
W W W
W
  
  
  
+ 
+ = +  + =

u v
u v
u v u v
Teorema 4.3.5 (Soma de subespaços) 
Sejam 1W e 2W subespaços de um espaço vetorial V. Então, o 
conjunto 1 2 1 2 2{ , e }W W W W+ = =  1 2 1v w + w w w é um 
subespaço de V. 
 
0
) 0
0
x y
a z t
x y z t
+ =

− =
 − − + =
2 1 2
1 1 1 1 0
1 1 0 0 0
0 0 1 1 0
L L L
− − 
 
→ − + 
 − 
1 1 1 1 0
0 2 1 1 0
0 0 1 1 0
− − 
 
− 
 − 
3 e 4p n= =
2 0
2 0 0
z t
y z t
y t t y
=
+ − =
+ − =  =
0
0 0
0
x y z t
x z z
x
− − + =
− − + =
=
1 2 {(0,0, , )}W W t t =  1 2 1 2) {(0,0,1,1)} ( ) 1W Wb Dim W W  =   =
1 1 0 0
0 0 1 1
1 1 0 0
1 0 1 0
1 0 0 1
− 
 
 
 
 
 
 − 
   
 
1
1
( , , , ) [( , ,0,0) (0,0, , )]
[(1, 1,0,0),(0,0,1,1)]
W x x t t x x t t
W
= − = − +
= −
   2 ( , , , ) ( , ,0,0) ( ,0, ,0) ( ,0,0, )W y z t y z t y y z z t t= + − = + + −
1 1 0 0
0 0 1 1
0 2 0 0
0 1 1 0
0 1 0 1
− 
 
 
 
 
 
 − 
1 1 0 0
0 1 0 0
0 0 1 1
0 1 1 0
0 1 0 1
− 
 
 
 
 
 
 − 
1 1 0 0
0 1 0 0
0 0 1 1
0 0 1 0
0 0 0 1
− 
 
 
 
 
 
 
 
1 1 0 0
0 1 0 0
0 0 1 1
0 0 0 1
0 0 0 1
− 
 
 
 
 
 
 
 
1 1 0 0
0 1 0 0
0 0 1 1
0 0 0 1
0 0 0 0
− 
 
 
 
 
 
 
 
 2 [(1,1,0,0),(1,0,1,0),( 1,0,0,1)]W = −
1 2
4 4
( ) 4
p e n
Dim W W
= =
+ =
d)
c) Não, pois a dimensão de Dim(W1∩W2) = 1
1 2 4?W W+ =
4
1 2 1 2
4
4
1 2
( ) 4, , 
( ) 4
Dim W W W W
Dim
W W
+ = 
=
+ =
0
0
0
0
a d
b c
a c
b d
− =

− =

− =
 − =
1 0 0 1 0
0 1 1 0 0
1 0 1 0 0
0 1 0 1 0
− 
 
− 
 −
 
− 
1 0 0 1 0
0 1 1 0 0
0 0 1 1 0
0 1 0 1 0
− 
 
− 
 −
 
− 
1 0 0 1 0
0 1 1 0 0
0 0 1 1 0
0 0 1 1 0
− 
 
− 
 −
 
− 
1 0 0 1 0
0 1 1 0 0
0 0 1 1 0
0 0 0 0 0
− 
 
− 
 −
 
 
1 0 0 1 0
0 1 0 1 0
0 0 1 1 0
0 0 0 0 0
− 
 
− 
 −
 
 
1 0 0 1 0
0 1 0 1 0
0 0 1 1 0
0 0 0 0 0
− 
 
− 
 −
 
 
0
0
0
c d c d
b d b d
a d a d
− =  =
− =  =
− =  =
{( , , , ), }S d d d d d=   1 2 {(1,1,1,1)}W W  =
1
0 0
0 0
1 0 0 1
 =
0 1 1 0
a b a b
W
b a a b
a b
        
= = +        
        
    
+    
    
2
0 0
0 0
1 0 0 1
 =
1 0 0 1
a b a b
W
a b a b
a b
        
= = +        
        
    
+    
    
1 2[W ] [W ] 2Dim Dim= = 
1 2 1 2 1 2[W W ] [W ] [W ] [W W ]Dim Dim Dim Dim + = + − 
1 2[W W ] 2 2 1 3Dim + = + − =  Não é soma direta
   1 ( , , 2 ) [( ,0, ) (0,, 2 )]
 = [ (1,0, 1) (0,1, 2)]
V x y x y x x y y
x y
= − − = − + −
− + −
1
{(1,0, 1), (0,1, 2)}V = − −  1dim( ) 2V =
Pelo o algoritmo do Espaço Linha
1
1
1 0 1
0 1 2
V
V
− 
 
− 
1 0 1
0 1 2
0 0 1
− 
 
− 
 
 
22
[(0,0,1)] {(0,0,1)}VV  =  =
1 2 1 2V V V V   = 0
1 0 1
0 1 2
− 
 
− 
1 1 0
0 0 1
 
+ 
 
1 0 1
0 1 2
1 1 0
0 0 1
− 
 
− 
 
 
 
1 0 1
0 1 2
0 1 1
0 0 1
− 
 
− 
 
 
  3 4
1 0 1
0 1 2
0 0 3
30 0 1 L L
− 
 
− 
 
 
= 
1 0 1
0 1 2
0 0 1
0 0 0
− 
 
− 
 
 
 
1 23 ( ) 3p Dim V V=  + =
Não é soma direta.
( ) 2, ( ) 2 e ( ) 1
( ) ( ) ( ) ( ) 2 2 1 3
Dim U Dim W Dim U W
Dim U W Dim U Dim W Dim U W
= =  =
+ = + −  = + − =
3[ ] {( , , ) / 2 0}
[ ] [( ,2 , )] [( ,2 ,0) (0,0, )] [ (1,2,0) (0,0,1)]
{(1,2,0),(0,0,1)} dim( ) 2U
U x y z x y
U x x z x x z x z
W
=  − =
= = + = +
=  =
3[ ] {( , , ) / 2 0}
[ ] [( 2 , , )] [( , ,0) ( 2 ,0, )] [ (1,1,0) ( 2,0,1)]
{(1,1,0),( 2,0,1)} dim( ) 2W
W x y z x y z
U y z y z y y z z y z
W
=  − + =
= − = + − = + −
= −  =
2 0 2 2
2 0 2 2 0 2 0
2
x y y x y x
U W
x y z x x z x z
x
z
− = = =  
 =    
− + = − + = − + =  
= ,2 , /
2
x
U W x x x
  
 =   
  
1
1,2,
2
U W 
  
=   
  
( 1,1) (1,0) (0,1)
1
1
a b
a
b
− = +
= −

=
(1,1) (1,0) (0,1)
1
1
c d
c
d
= +
=

=
  1
1 1
1 1
a c
I
b d


−   
 = =   
   
 
1
1 1
2 2
1 1
2 2
a c
I
b d


 
−  
 = =   
  
 
 (1,0) ( 1,1) (1,1)
1 1
2 1
0 2
a b
a b
b b
a b
= − +
− + =
 =  =
+ =
(0,1) ( 1,1) (1,1)
0 1
2 1
1 2
c d
c d
d d
c d
= − +
− + =
 =  =
+ =
1
0
2
c d c d− + =  = =
1
0
2
a b a b+ =  = − = −
 
2
3 1
6 2
3 1
6 2
a c
I
b d


 
 
 
  = = 
  
− 
 
(1,0) ( 3,1) ( 3, 1)
1 33 3 1
2 3 1
62 30 3 3 1
3
6
a b
a ba b
b b
a b b b
a
= + −
=  +  = 
  =  = = 
− = + =  
=
(0,1) ( 3,1) ( 3, 1)
3 3 0 3 13 3 0 3
6 3
3 3 3 6 21 3
c d
c dc d
c c
c dc d
= + −
 + = +  = 
  =  = = 
− =− =  
1
1
1 1
1
2 2
c d
d c
d
− =
= −
= − = −
(1,0) (2,0) (0,2)
1
2 1
2
2 0 0
a b
a a
b b
= +

=  =

 =  =
(0,1) (2,0) (0,2)
2 0 0
1
2 1
2
c d
c c
d d
= +
=  =


=  =

 
3
1
0
2
1
0
2
a c
I
b d


 
  
 = =   
  
 
 
) (3, 2) (1,0) (0,1)
3
2
i a b
a
b
− = +
=

= −
 
3
(3, 2)
2
 
 − =  
− 
) (3, 2) ( 1,1) (1,1)
3 1
2 1
2 2
3 5
2 5
2 2
ii a b
a b
b b
a b
a b
a a
a b
− = − +
− + =
 =  =
+ = −
− = −
 = −  = −
+ = −
 
1
5
2
(3, 2)
1
2

 
− 
 − =  
 
 
 
(3, 2) ( 3,1) ( 3, 1)
3 3 3
2
3 3
2 1 3 2 3 3 2 3 3
1
2 3 2 3 2 33 3
1 1
a b
a b
a b
a
− = + −
  +  =

− = −
− − − + −
= = = = −
−
−
3 3
1 2 2 3 3 3 2 3 3
1
2 3 2 3 2 33 3
1 1
b
− − − +
= = = = +
−
−
2 {( 3,1),( 3, 1)} = −
iii) 2
 
3
3 2 3
2 3
(3, 2)
3 2 3
2 3

 −
 
 − =
 +
  
 
(3, 2) (2,0) (0,2)
3
2 3
2
2 2 1
a b
a a
b b
− = +

=  =

 = −  = −
 
3
3
(3, 2) 2
1

 
  − =
 
− 
IV)
( , ) 4( 1,1) 0(1,1) ( 4,4)x y = − +  = −v
 
4
i) ( 4,4) (1,0) (0,1)
4
a b

− 
− = +  =  
 
v
ii) ( 4,4) ( 3,1) ( 3, 1)a b− = + −
3 3 4 ( 3)
4 (3)
a b
a b
  +  = − 

− = 
3 3 4 3
3 3 12
6 12 4 3
6 2 3
3
a b
a b
a
a
 + = −

− =
= −
−
=
3 3 4 3
3 3 12
6 12 4 3
6 2 3
3
a b
a b
b
b
− − =

− =
− = +
− −
=
 
2
6 2 3
3
6 2 3
3

 −
 
  =
 − −
 
 
v
 
3
) ( 4,4) (2,0) (0,2)
2 4 2
2 4 2
2
2
iii a b
a a
b b

− = +
= − = − 
 
= = 
− 
 =  
 
v
4 2 0
Note : 2 2
4 0 2
−     
= − +     
     
         ( )  
'
, '
1
'
I I
 
     
−
=   = v v v v
     
'
'
)a I

  
= v v
 
1 1 0 1 1
0 1 1 2 1
1 0 1 3 4

−    
    
= − =    
    − −    
v
   ( )  
1
'
'
) b I

  
−
= v v
  '
1 1 1
2 2 2 1 1
1 1 1
1 2
2 2 2
4 3
1 1 1
2 2 2

 
 
−    
    = − − =    
   −    
 − 
 
v
1 1 1 0
1 0 1 1
4 1 0 1
a
b
c
    
    
= −    
    − −    
1
1
4
2 2
1
a b
b c
a c
a
a
+ =

− + =
 − = −
= −
= −
'
1
[ ] 2
3

− 
 
=  
 
 
v
      ,
'
Ou
I

  
= v v
4
4 1
3
1
3 1
2
a c
c
c
b c
b
b
− = −
− =
=
− + =
− =
=
1 3cos
3 3 2 2
3 3 1
cos
3 3 2 2
sen
R
sen
 

 
      
− − −      
        − = =          
− − −      
      
{(1,0),(0,1)} =
 
'
1 3
2 2
) 
3 1
2 2
a I


 
− 
 =
 
 
 
   ( )
1
1
'
'
3
1 3 1 3 1 3
2 2 2 2 2 2
) 
3 1 3 1 3 1
2 2 2 2 2 2
 1
b I I R
 
 
−
−
 
 
 
     
−     
     = = = = =
     
− −     
     
)
( 1,0) (1,0) (0,2)
1
2 0 0
i
a b
a
b b
− = +
= −

=  =
(1,1) (1,0) (0,2)
1
1
2 1
2
c d
c
d d
= +
=


=  =
  2
1
1 1
1
0
2
I


− 
  =
 
 
)
( 1, 1) ( 1,0) (1,1)
1
1
1 1
1 1 0
ii
a b
a b
b
a
a a
− − = − +
− + = −

= −
− − = −
− + =  =
(0, 1) ( 1,0) (1,1)
0
1
0
1
c d
c d
d
c d
d c
− = − +
− + =

= −
− + =
= = −
  3
2
0 1
1 1
I


− 
 =  
− − 
)
( 1, 1) (1,0) (0,2)
1
1
2 1
2
iii
a b
a
b b
− − = +
= −


= −  = −
(0, 1) (1,0) (0,2)
0
1
2 1
2
c d
c
d d
− = +
=


= −  = −
  3
1
1 0
1 1
2 2
I


− 
  =
 − −
 
     2 3 3
1 2 1
1 1 1 0
0 1
) . 1 1 1
1 10
2 2 2
iv I I I
  
  
− −   
−    = = =    − − − − 
   
     2 3 3
1 2 1
) .b I I I
  
  
=
1 0 1 0 0 1 0 0
0 0 0 0 0 0 0 1
a b c
       
= + +       
       
1 1 1 0 0 1 0 0
0 0 0 0 0 0 0 1
d e f
       
= + +       
       
1 1 1 0 0 1 0 0
0 1 0 0 0 0 0 1
g h i
       
= + +       
       
1, 0 e 0a b c= = =
1, 1 e 0d e f= = =
1, 1 e 0g h i= = =
  1
1 1 1
0 1 1
0 0 1
I


 
 
=  
 
 
  1
a d g
I b e h
c f i


 
 
=  
 
 
{(1,2),(2,1)} =
(1,2) (1,2) (2,1)
2 1
2 2
2 4 2
2 2
3 0
0
2 2
2 0 2
1
a b
a b
a b
a b
a b
b
b
a b
a
a
= +
+ =

+ =
− − = −

+ =
− =
=
+ =
+ =
=
(2,1) (1,2) (2,1)
2 2
2 1
2 4 4
2 1
3 3
3
1
3
2 1
2 1 1
2 0
0
a b
a b
a b
c d
c d
d
d
c d
c
c
c
= +
+ =

+ =
− − = −

+ =
− = −
−
= =
−
+ =
+ =
=
=
 
1 0
0 1
I


 
=  
 
(1,4)
(4,1)
1 4 3
1
4 1 3
4 1( 1)
4 1
5 0
A
B
a
y x
y x
x y
− −
= = = −
−
− = − −
− = − +
+ − =
Equação vetorial de uma reta
(4,1) (1,4) (3, 3)
(1,4)
( ) (1,4) (3, 3)
1 3
4 3
AB B A
A
f t A ABt t
x t
y t
= − = − = −
= + = + −
= +

= −
5x y+ =
1
1 4 1 0
4 1 1
4 4 1 16 0
3 3 15 0
5 0
x y
x y x y
x y
x y
=
+ + − − − =
+ − =
+ − =