Post on 11-Oct-2015
5/20/2018 Ad1-2014 Algebra Linear
1/10
a)Espao da Coluna de A e B:
C(A)={(1,1,3), (1,3,1), (0,1,-1)}
L2-L1 L3-(3xL1)
L3-(-L2)
Colunas Pivot
Base {(1,0,0), (0,1,0)} Dim R = 2 (DIMENSO 2)
C(B)={(1,0,0), (1,2,0), (0,1,0)} =
1 1 0
1 3 1
3 1 -1
1 1 0
0 2 1
3 1 -1
1 1 0
0 2 1
0 -2 -1
1 1 0
0 2 1
0 0 0
5/20/2018 Ad1-2014 Algebra Linear
2/10
Colunas Pivot
Base {(1,0,0), (0,1,0)} Dim R = 2 (DIMENSO 2)
B)Espao da Linha de A e B:
L(A)={(1,1,0), (1,3,1), (3,1,-1)}
L2-L1 L3-(3xL1)
L3-(-L2)
Base {(1,1,0), (0,2,1)} Dim R = 2 (DIMENSO 2)
L(B)={(1,1,0), (0,2,1), (0,0,0)}
Base {(1,1,0), (0,2,1)} Dim R = 2 (DIMENSO 2)
c) Os espaos da colunas de A e de B so iguais e Os espaos da LINHA de A e de B so iguais.
1 1 0
0 2 1
0 0 0
1 1 0
1 3 1
3 1 -1
1 1 0
0 2 1
3 1 -1
1 1 0
0 2 1
0 -2 -1
1 1 0
0 2 1
0 0 0
1 1 0
0 2 1
0 0 0
5/20/2018 Ad1-2014 Algebra Linear
3/10
C(A){(1,0,0), (0,1,0)} = C(B) {(1,0,0), (0,1,0)}
L(A){(1,1,0), (0,2,1)}= L(B){(1,1,0), (0,2,1)}
C1(A)=(1,1,3)
L2(A)=(1,3,1)
d(C1(A), L2(A)) = | (1,1,3)- (1,3,1)|
d(C1(A), L2(A)) =
d(C1(A), L2(A)) =
d(C1(A), L2(A)) =
Logo d(C1(A), L2(A)) =
Cos = C1(A) . L2(A)
|C1(A)| . |L2(A)|
C1(A) . L2(A) = (1,1,3) . (1,3,1)
C1(A) . L2(A) = (1.1 + 1.3 + 3.1)= 1 + 3 + 3 =
C1(A) . L2(A) = 7 (Produto Interno)
|C1(A)| = = =
|L2(A)| = = = =
Cos = 7 = 7 = 7 = 0,63 51
. 11
5/20/2018 Ad1-2014 Algebra Linear
4/10
proj(C1(A),C2(A))L2(A)= projC1(A)L2(A)+projC2(A)L2(A)
Sendo assim:
proj(C1(A),C2(A))L2(A)= proj(1,1,3)(1,3,1)+proj(1,3,1)(1,3,1)=
=((1,3,1).(1,1,3)). (1,1,3) + ((1,3,1).(1,3,1)). (1,3,1) =1 + 1 + 3 1 + 3 + 1
= 1+3+3 . (1,1,3) + 1+9+1 .(1,3,1) = 7(1,1,3) + 11(1,3,1)=11 11 11 11
= (7, 7, 21) + (11, 33, 11) = 18 , 40, 3211 11 11 11 11
Logo:proj(C1(A),C2(A))L2(A)= 18 , 40 , 32
11 11 11
Conforme verificado no exerccio 1 na letra a base gerada foi:
Base {(1,0,0), (0,1,0)} Dim R = 2
(1,0,0) . (0,1,0) = 0+0+0 = 0Uma base ortogonal quando dois a dois vetores so ortogonais, se o produto interno deles = 0
5/20/2018 Ad1-2014 Algebra Linear
5/10
X1= 2X4
X1 X2 X3 X4
2X4 Incgnita Incgnita Incgnita
(X1,X2,X3, X4) =(2X4, X2, X3, X4)
Temos ento a Dimenso 3
(X1,X2,X3, X4) =X1( 0,0,0,0) + X2( 0,1,0,0) + X3( 0,0,1,0) + X4( 2,0,0,1)
(X1,X2,X3, X4) =X2( 0,1,0,0) + X3( 0,0,1,0) + X4( 2,0,0,1)
B={ ( 0,1,0,0), ( 0,0,1,0), ( 2,0,0,1)}
u e v so Linearmente Independente se:
u v = 0
Ou seja:
a( 0,1,0,0)+b(0,0,1,0) +c(2,0,0,1)=
2c=0
a=0b=0
c=0
Logo = ( 0,1,0,0), ( 0,0,1,0), ( 2,0,0,1) so LI
X1+ X2+ X3= 0
X1 = -X2- X3
X3+ X4= 0
X3 = -X4
X1 X2 X3 X4
-X2-X3 Incgnita -X4 Incgnita
(X1,X2,X3, X4) =(-X2-X3, X2, -X4, X4)
5/20/2018 Ad1-2014 Algebra Linear
6/10
Temos ento a Dimenso 3 pois:
(X1,X2,X3, X4) =X1( 0,0,0,0) + X2( -1,1,0,0) + X3( -1,0,0,0) + X4( 0,0,-1,1)
(X1,X2,X3, X4) =X2( -1,1,0,0) + X3( -1,0,0,0) + X4( 0,0,-1,1)
B={( -1,1,0,0) ,( -1,0,0,0) ,( 0,0,-1,1)}
u e v so Linearmente Independente se:
u v = 0
Ou seja:
a( -1,1,0,0)+b( -1,0,0,0) +c( 0,0,-1,1)=
-a-b=0
a=0
-c=0c=0
Logo =( -1,1,0,0) ,( -1,0,0,0) ,( 0,0,-1,1) so LI
(1,1,1,1),(1,2,3,4),(2,3,4,5)
Sejam a; b; Consideremos
a(1,1,1,1) + b(1,2,3,4) + c(2,3,4,5)= (a+b+2c, a+2b+3c, a+3b+4c, a+4b+5c) = (x,y,z,w)
a+ b+ 2c=x (I)
a+2b+3c=y (II)
a+3b+4c=z (III)
a+4b+5c=w (IV)
L2L2-L1 (II)a+2b+3c=y
-a -b -2c =-x
b+c=-x+y
L4L4-L1 (IV)a+4b+5c=w
-a -b -2c =-x
3b+3c=-x+w
L3L3-L1 (III)
a+3b+4c=z
-a -b -2c =-x
2b+2c=-x+z
Novo Sistema:
a+ b+ 2c=x (I)b+c=-x+y (II)
2b+2c=-x+z (III)3b+3c=-x+w (IV)
5/20/2018 Ad1-2014 Algebra Linear
7/10
Sistema II:
a+ b+ 2c=x (I)
b+c=-x+y (II)2b+2c=-x+z (III)3b+3c=-x+w (IV)
L3L3-2L2 (III)
2b+2c=-x+z-2b-2c=+2x-2y
0+0=x-2y+z
L4L4-3L2 (III)
3b+3c=-x+w-3b-3c= 3x-3y
0+0=2x-3y+w
Sistema III:
a+ b+ 2c=x (I)b+c=-x+y (II)0+0=x-2y+z(III)0+0=2x-3y+w(IV)
Logo vimos por L4 que x=3y-w e em L3 que x=2y-z.Igualando-as temos:2
3y-w=2y-z2
3y-w=4y-2z
2z-w=4y-3y
2z-w=y
-y+2z-w=x
x=-y+2z-w
x=-y+2z-w. Logo S={(x,y,w,z) 4| x=-y+2z-w} ou seja (-y+2z-w,y,z,w)
Temos ento a Dimenso 3 pois:
(x,y,z,w ) =x( 0,0,0,0) + y( -1,1,0,0) + z( 2,0,1,0) + w( -1,0,0,1)
(x,y,z,w ) =y( -1,1,0,0) + z( 2,0,1,0) + w( -1,0,0,1)
B={ ( -1,1,0,0),( 2,0,1,0),( -1,0,0,1) }
u e v so Linearmente Independente se:
u v = 0
5/20/2018 Ad1-2014 Algebra Linear
8/10
Ou seja:
a( -1,1,0,0)+b(2,0,1,0) +c(-1,0,0,1)=
-a+2b-c=0
a=0
b=0c=0
Logo = ( -1,1,0,0),( 2,0,1,0),( -1,0,0,1) so LI
Tambm podemos olhar no sistema III novamente e enxergar de outra forma:
Sistema III:
a+ b+ 2c=x (I)b+c=-x+y (II)
0+0=x-2y+z(III)0+0=2x-3y+w(IV)
Em L3 z=-x+2y e em L4 que w=-2x+3y.Igualando-as temos:
Logo S={(x,y,w,z) 4| z=-x+2y;w=-2x+3y} ou seja (x,y,-x+2y,-2x+3y)
Temos ento a Dimenso 2 pois:
(x,y,z,w ) =x( 1,0,-1,-2) + y( 0,1,2,3) + z( 0,0,0,0) + w( 0,0,0,0)
(x,y,z,w ) =x( 1,0,-1,-2) + y( 0,1,2,3)
B={ ( 1,0,-1,-2), ( 0,1,2,3) }
u e v so Linearmente Independente se:
u v = 0
Ou seja:
a( 1,0,-1,-2)+b( 0,1,2,3) =
a=0
b=0
-a+2b=0
-2a+3b=0
Logo ( 1,0,-1,-2),( 0,1,2,3) so LI
5/20/2018 Ad1-2014 Algebra Linear
9/10
Uma base ortogonal quando dois a dois vetores so ortogonais, se o produtointerno deles = 0.
Dizemos que v e w so ortogonais se = 0.
Por exemplo v=(1, 0, 1, 0, 1, 0), w= (0, 1, 0, 1, 0, 1)
Soluo: De fato, (1, 0, 1, 0, 1, 0) .(0, 1, 0, 1, 0, 1) = 1(0)+0(1)+1(0)+0(1)+1(0)+0(1) =0
O vetor nulo o mesmo em W e V, isto , V=W=, assim VW no vazio.
Exemplo: Sejam V e W subespaos vetoriais de R, definidos por:
V= = {(x,y,0): xR, yR }W= = {(0,0,z): zR }
O conjunto VW um subespao de R e observamos que V W ={} o subespao nulo.
O ngulo de 90 logo a base ortogonal.
Dois vetores, ve w, pertencentes ao espao vetorial Vcom produto interno, so vetores
ortogonaisse, e somente se, seu produto interno nulo, ou seja, w,v=0.
X1+ 2X2-2X3= b1 (I)
2X1+ 5X2-4X3= b2 (II)
4X1+ 9X2-8X3= b3(III)
(II) L2L2-(2L1)
2X1+ 5X2-4X3= b2-2X1- 4X2+4X3= - 2b1
X
2
= - 2
1
2
1 2 -2 b1
2 5 -4 b24 9 -5 b3
1 2 -2 b1
0 1 0 -2b1+b2
4 9 -5 b3
5/20/2018 Ad1-2014 Algebra Linear
10/10
(III) L3L3-(4L1)
4X1+ 9X2-8X3= b3-4X1- 8X2+8X3= - 4b1
X
2
= - 4
1
3
Ento temos:
X1+ 2X2-2X3= b1 (I)
X2 = - 2b1 +b2 (II)X2 = - 4b1 +b3(III)
Por L3 temos que X2 = - 4b1 +b3e por L2 temos
que X2 = - 2b1 +b2 .
Igualando esses dois valore de X2temos:
- 2 1 + 2 = - 4 1 + 3 == - 2 1 + 4 1= - 2 + 3 == 2 1= - 2 + 3
1= - 2 + 32
Logo : 1= - 2 + 32
Sendo assim este um Sistema Possvel Indeterminado (SPI)
1 2 -2 b1
0 1 0 -2b1 +b2
0 1 0 -4b1+b3
L3L3-L2
1 2 -2 b1
0 1 0 -2b1 +b2
0 0 0 2b1-b2+ b3
Sem < n o sistema nunca pode
ter soluo nica.
Ou seja:
Sistema Possvel Indeterminado