Álgebra Linear Espaços e subespaços vetoriais - exercícios explicados 2

Prévia do material em texto

1 
 
 
1) Seja S o subespaço de 4R definido por: 
( ) 4, , , / 2 0 0S x y z w R x y z t=  + − =  = 
Pergunta-se: 
a) ( 1, 2,3,0) ?S−  
SOLUÇÃO: 
( )
1; 2; 3; 0
2 1 2.2 3 1 4 3 0 0
1,2,3,0
x y z t
x y z t
S
= − = = =
+ − = − + − = − + − =  =
− 
 
b) (3,1,4,0) ?S 
SOLUÇÃO: 
( )
3; 1; 4; 0
2 3 2.1 4 3 2 4 1 0
3,1,4,0
x y z t
x y z
S
= = = =
+ − = + − = + − = 

 
c) ( 1,1,1,1) ?S−  
SOLUÇÃO: 
( )
1; 1; 1; 1
1 0
1,1,1,1
x y z t
t
S
= − = = =
= 
− 
 
2) Seja S o subespaço de (2,2)M : 
2
/ ,
a b a
S a b R
a b b
 −  
=   
+ −  
 
Pergunta-se: 
a) 
5 6
1 2
S
 
 
 
? 
SOLUÇÃO: 
2 6 3
2 2
3 ( 2) 3 2 5( )
3 ( 2) 3 2 1( )
a a
b b
a b V
a b V
=  =
− =  = −
− = − − = + =
+ = + − = − =
 
2 
 
5 6
1 2
S
 
 
 
 
b) Qual deve ser o valor de k para que o vetor 
4
?
2 3
k
S
− 
 
− 
 
SOLUÇÃO: 
( )
4
2 2 1 2
2 3 2 1
3 3
2
a b
a k k
a b a a
b b
k
− = −
=  = − = −
+ =  + =  = −
− = −  =
= −
 
3) Determinar os subespaços de 3R gerados pelos seguintes conjuntos: 
a) ( ) 2, 1,3A = − 
SOLUÇÃO: 
Seja 
( )
( ) ( )
( ) 
, ,
/ , , 2, 1,3
2 2
3 3
2 , , 3 /
x y z A
a R x y z a
x a x y
y a a y
z a z y
A y y y y R

  = −
=  = −

= −  = −
 =  = −
= − − 
 
b) ( ) ( ) 1,3,2 , 2, 2,1A = − − 
SOLUÇÃO: 
Seja ( ), ,x y z A , ( ) ( ) ( ) ( ), / , , 1,3,2 2, 2,1 2 ,3 2 ,2a b R x y z a b a b a b a b  = − + − = − + − + 
( ) 
2 ( )
3 2 ( )
2 ( )
( ) ( ) 2 3 2 2
3
3( ) ( ) 3 3 6 3 2 4
4
3
2 4 4 4 3 7 5 4 0
4
, , / 7 5 4 0
x a b I
y a b II
z a b III
I II x y a b a b a
x y
I II x y a b a b b b
x y
z a b x y z x y x y x y z
A x y z x y z
= − +

= −
 = +
+  + = − + + − =
+
+  + = − + + − =  =
+
= + = + +  = + + +  + − =
= + − =
 
 
 
 
 
 
 
 
 
 
 
 
 
 
3 
 
c) ( ) ( ) ( ) 1,0,1 , 0,1,1 , 1,1,0A = − 
SOLUÇÃO: 
Seja ( ), ,x y z A 
( ) ( ) ( ) ( ) ( )
( ) 
, , / , , 1,0,1 0,1,1 1,1,0 , ,
( )
( )
( )
( ) ( )
, , /
a b c R x y z a b c a c b c a b
x a c I
y b c II
z a b III
I II x y a c b c a b
z a b z x y
A x y z z x y
  = + + − = − + +
= −

= +
 = +
+  + = − + + = +
= +  = +
= = +
 
d) ( ) ( ) ( ) 1,0,1 , 0,1,1 , 1,1,0A = − 
SOLUÇÃO: 
Seja ( ), ,x y z A 
( ) ( ) ( ) ( ) ( ), , / , , 1,1,0 0,1, 2 2,3,1 2 , 3 , 2
2 ( )
3 ( )
2 ( )
a b c R x y z a b c a c a b c b c
x a c I
y a b c II
z b c III
  = − + − + − = − − + + − +
= − −

= + +
 = − +
 
Não existe nenhuma relação entre , ,x y z , logo, 3A R= 
e) ( ) ( ) ( ) ( ) 1,2, 1 , 1,1,0 , 3,0,1 , 2, 1,1A = − − − − − 
SOLUÇÃO: 
Seja ( ), ,x y z A 
( ) ( ) ( ) ( ) ( )
( ) ( )
, , , / , , 1, 2, 1 1,1,0 3,0,1 2, 1,1
, , 3 2 ,2 ,
3 2 ( )
2 ( )
( )
( ) ( ) 3( ) 3 3 2 2 3 3 3 0
a b c d R x y z a b c d
x y z a b c d a b d a c d
x a b c d I
y a b d II
z a c d III
I II III x y z a b c d a b d a c d
  = − + − + − + − −
= − − − + − − + +
= − − −

= + −
 = − + +
+ +  + + = − − − + + − − + + =
 
( ) 3, , / 3 0A x y z R x y z=  + + = 
f) ( ) ( ) ( ) ( ) 1,2, 1 , 1,1,0 , 0,0,2 , 2,1,0A = − − − 
SOLUÇÃO: 
Seja ( ), ,x y z A 
( ) ( ) ( ) ( ) ( )
( ) ( )
, , , / , , 1, 2, 1 1,1,0 0,0,2 2,1,0
, , 2 ,2 , 2
2 ( )
2 ( )
2 ( )
a b c d R x y z a b c d
x y z a b d a b d a c
x a b d I
y a b d II
z a c III
  = − + − + + −
= − − + + − +
= − −

= + +
 = − +
 
Como o escalar c só está presente na equação (III) não é possível estabelecer uma relação 
entre , ,x y z , logo, 3A R= . 
 
 
 
4 
 
4) Seja o conjunto  1 2,A v v= , sendo ( ) ( )1 21,3, 1 , 1, 2,4v v= − − = − . 
Determinar: 
a) O subespaço ( )G A . 
SOLUÇÃO: 
( ) ( ) 1,3, 1 , 1, 2,4A = − − − 
Seja ( ), ,x y z A 
( ) ( ) ( )
( ) ( )
( )
, / , , 1,3, 1 1, 2,4
, , ,3 2 , 4
( ) 2 3
3 2 ( ) 3 2 3 2 2 2 2
4 ( )
a b R x y z a b
x y z a b a b a b
x a b I b x a b x y x x y
y a b II y a x a y a x a y a x a y x
z a b III
  = − − + −
= − + − − +
= − +  = +  = + + = +

= −  = − +  = − −  = −  = +

= − +
 
( ) ( )
( ) 3
2 4 3 2 12 4
10 3 10 3 0
, , /10 3 0
z y x x y z y x x y
z x y x y z
A x y z R x y z
= − + + +  = − − + +
= +  + − =
=  + − =
 
b) O valor de k para que o vetor ( )5, ,11v k= pertença a ( )G A . 
SOLUÇÃO: 
Pelo item a o vetor ( )5, ,11v k= pertença a ( )G A se: 
5, , 11
10.5 3 11 0 50 3 11 0 3 39
13
x y k z
k k k
k
= = =
+ − =  + − =  = −
= −
 
5) Sejam os vetores ( ) ( ) ( )1 2 31,1,1 , 1,2,0 , 1,3, 1v v v= = = − . Se ( )  1 2 33, 1, , ,k v v v−  , qual o valor 
de k ? 
SOLUÇÃO: 
( ) ( ) ( ) ( )3, 1, 1,1,1 , 1,2,0 , 1,3, 1k−  −   
( ) ( ) ( ) ( )
( ) ( )
, , / 3, 1, 1,1,1 1,2,0 1,3, 1
3, 1, , 2 3 ,
3 ( )
1 2 3 ( )
( )
( 2)( ) ( ) 6 1 2 2 2 2 3 7
7 7
a b c R k a b c
k a b c a b c a c
a b c I
a b c II
k a c III
I II a b c a b c a c
a c k a c
  − = + + −
− = + + + + −
= + +

− = + +
 = −
− +  − − = − − − + + +  − + = −
− =  = − =
 
6) Determinar os subespaços de 2P (espaço vetorial dos polinômios de grau 2 ) 
gerados pelos seguintes vetores: 
a) 2 2
1 2 32 2, 3, 2p x p x x p x x= + = − + + = + 
 
 
 
 
 
 
 
 
 
 
5 
 
SOLUÇÃO: 
( ) ( ) ( )
( ) ( ) ( )
( )
2
2 2
2
2
2 2 2
2 2 2
2 2
( ) 2 2, 3, 2
( )
, , / 2 2 3 2
2 2 3 2
2 2 2 3
2 2 3 2 3 2 2 2
p ax bx c
G P x x x x x
p G P
k l m R ax bx c k x l x x m x x
ax bx c kx k lx lx l mx mx
ax bx c l m x k l m x k l
l m a m a l
k l m b c l l a l b c l l a l b c
= + +
 = + − + + + 

  + + = + + − + + + +
+ + = + − + + + +
+ + = − + + + + +
− + =  = +
+ + =  − + + + =  − + + + =  +
2 3 2 3
2 0
a b
k l c k c l
a b c


=

+ =  = −
− + =
 
b) 2 2
1 2,p x p x x= = + 
SOLUÇÃO: 
2
2 2
2
2
( ) ,
( )
p ax bx c
G P x x x
p G P
= + +
 = + 

 
( ) ( )
( )
2 2 2
2 2 2
2 2
, /
0
k l R ax bx c k x l x x
ax bx c kx lx lx
ax bx c k l x lx
k l a
l b
c
  + + = + +
+ + = + +
+ + = + +
+ =

=
 =
 
 22( ) / ,G P ax bx a b R= +  
c) 21 2 31, ,p p x p x= = = 
SOLUÇÃO: 
2
2
2
2
( ) 1, ,
( )
p ax bx c
G P x x
p G P
= + +
 =  

 
( ) ( ) ( )2 2
2 2
, , / 1k l m R ax bx c k l x m x
ax bx c k lx mx
m a
l b
k c
  + + = + +
+ + = + +
=

=
 =
 
2 2( )G P P= 
 
 
 
 
6 
 
7) Determinar o subespaço ( )G A para ( ) ( ) 1, 2 , 2,4A = − − . O que representa 
geometricamente este subespaço? 
SOLUÇÃO: 
( ) ( ) 1, 2 , 2,4A = − − 
Seja ( ), ( )x y G A 
( ) ( ) ( )
( ) ( )
( )
( ) 
, / , 1, 2 2,4
, 2 , 2 4
2 ( ) 2
2 4 ( ) 2 2 4 2 4 4
2
( ) , 2 /
a b R x y a b
x y a b a b
x a b I a x b
y a b II y x b b y x b b
y x
G A x x x R
  = − + −
= − − +
= −  = +

= − +  = − + +  = − − +
= −
= − 
 
( ) ( )0,0 , 1, 2 ( )O B G A= = −  , geometricamente este subespaço é uma reta no plano 
cartesiano que passa pela origem. 
 
8) Mostrar que os vetores ( ) ( )1 22,1 , 1,1v v= = geram o 
2R . 
SOLUÇÃO: 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( )
( )
( ) ( ) ( )( ) ( )( )
2
2
, , / , 2,1 1,1
, 2 , , , 2 ,
2 2 2 2 2 2
2 2
, / , 2,1 2 1,1
x y R a b R x y a b
x y a a b b x y a b a b
x a b x y b b x y b b x y b b x y
y a b a y b a y x y a y x y a x y
x y R x y x y x y
    = +
= +  = + +
= +  = − +  = − +  − = −  = − +

= +  = −  = − − +  = + −  = −
  = − + − +
 
9) Mostrar que os vetores ( ) ( ) ( )1 2 31,1,1 , 0,1,1 , 0,0,1v v v= = = geram o 
3R . 
SOLUÇÃO: 
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( )( )
3
3
, , , , / , , 1,1,1 0,1,1 0,0,1
, , , 0, , 0,0, , , ,
, , / , , 1,1,1 0,1,1 0,0,1
x y z R a b c R x y z a b c
x y a a a b b c x y a a b a c
x a
y a b y x b b x y
z a c z x c c x z
x y z R x y z x x y x z
    = + +
= + +  = + +
=

= +  = +  = − +
 = +  = +  = − +
  = + − + + − +
 
 
 
 
 
7 
 
10) Seja o espaço vetorial (2,2)M . Determinar os subespaços gerados pelos vetores:a) 1 2
1 2 2 1
,
1 0 1 1
v v
−   
= =   
− −   
 
SOLUÇÃO: 
 1 2,
1 2 2 1
, /
1 0 1 1
2 2
0
2 2
x y
v v v
z w
x y
a b R a b
z w
x y a a b b
z w a b b
x y a b a b
z w a b b
 
=  
 
−     
  = +     
− −     
−     
= +     
− −     
− + +   
=   
− −   
 
( ) ( )
( )
2 2 2
2 2 2 2 2 3
2 3
(2,2)
x a b x z w w z w w z w
y a b y z w w y z w w z w
z a b z a w a z w
w b b w
z w z w
v M
z w
= − +  = − − + − = − + − = − −

= +  = − −  = − − = −

= −  = +  = −
 = −  = −
− − − 
=  
 
 
b) 
 1 2 3, ,
1 0 1 1 0 1
, , /
0 1 0 0 1 0
x y
v v v v
z w
x y
a b c R a b c
z w
 
=  
 
− −       
  = + +       
       
 
0 0
0 0 0 0
0
(2,2) / 0
x y a b b c
z w a c
x y a b b c
z w c a
x a b x w y z
y b c y b z b y z
z c
w a a w
x w y z x y z w
x y
v M x y z w
z w
− −       
= + +       
       
− + − +   
=   
   
= − +  = − − +

= − +  = − +  = − +

=
 =  =
= − − +  + − + =
 
=  + − + = 
 
 
11) Determinar os subespaços de 3P (espaço vetorial dos polinômios de grau 3 ) 
gerados pelos vetores: 
3 2 3 2
1 22 3, 2 3 2p x x x p x x x= + − + = − − + + 
 
 
 
 
 
 
8 
 
SOLUÇÃO: 
( ) ( )
( ) ( ) ( )
3 2
3 2 3 2
3
3
3 2 3 2 3 2
3 2 3 2 3 2
3 2 3 2
( ) 2 3, 2 3 2
( )
, / 2 3 2 3 2
2 3 2 3 2
2 2 3 3 2
2 2 2
p ax bx cx d
G P x x x x x x
p G P
k l R ax bx cx d k x x x l x x x
ax bx cx d kx kx kx k lx lx lx l
ax bx cx d k l x k l x k l x k l
a k l k a l k a
= + + +
 = + − + − − + + 

  + + + = + − + + − − + +
+ + + = + − + − − + +
+ + + = − + − + − + + +
= −  = +  = + ( )
( ) ( )
( )
( ) ( )
 3 23
2 2 3 2
2 2 3 2 6 4 5 3
3 2 3
3 2 3 3 2 2 9 6 2 2 11 8
5 3 11 8
( ) / 5 3 11 8
a c k a a c k a c
b k l b a c a c b a c a c b a c
c k l c a l l c a l l a c
d k l d a c a c d a c a c d a c
b a c d a c
G P ax bx cx d b a c d a c
+  = + +  = +

= −  = + − +  = + − −  = +

= − +  = − + +  = − +  = +
 = +  = + + +  = + + +  = +
= +  = +
= + + + = +  = +
 
12) Determinar o subespaço de 4R gerado pelos vetores 
( ) ( ) ( )2, 1,1,4 , 3,3, 3,6 , 0,4, 4,0u v w= − = − = − . 
SOLUÇÃO: 
Seja  , ,X u v w . 
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( )
, , , / , , /
, , , 2, 1,1,4 3,3, 3,6 0,4, 4,0
, , , 2 , , , 4 3 ,3 , 3 ,6 0,4 , 4 ,0
, , , 2 3 , 3 4 , 3 4 ,4 6
2 3
3 4 3 4 0
3 4
4 6
V x y z t a b c R V au bv cw
x y z t a b c
x y z t a a a a b b b b c c
x y z t a b a b c a b c a b
a b x
a b c y a b c y z y y z
a b c z
a b t
=   = + +
= − + − + −
= − + − + −
= + − + + − − +
+ =
− + + =  − − − =  − =  + =
− − =
+ =  ( )2 2 3 2a b t x t





 + =  =
 
( ), , , / 2 0V x y z t t x y z= =  + = . 
13) Verificar se o vetor ( )1, 3,2,0v = − − pertence ao subespaço do 4R gerado pelos 
vetores ( ) ( ) ( )1 2 32, 1,3,0 , 1,0,1,0 , 0,1, 1,0v v v= − = = − . 
SOLUÇÃO: 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
1, 3,2,0 2, 1,3,0 1,0,1,0 0,1, 1,0
1, 3,2,0 2 , ,3 ,0 ,0, ,0 0, , ,0
1, 3,2,0 2 , ,3 ,0
2 1 1
3 3 3 0
2 2 1 2 3
1, 3,2,0 0 2, 1,3,0 1 1,0,1,0 3 0,1, 1,0
1,
a b c
a a a b b c c
a b a c a b c
a b b
a c a a
a b c c c
− − = − + + −
− − = − + + −
− − = + − + + −
+ = −  = −

− + = −  − − = −  =
 + − =  − − =  = −
− − = − − − −
−( )  1 2 33,2,0 , ,v v v− 
 
 
os vetores 1 2 3; ;v v v .