Ad1-2014 Algebra Linear

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


2/10

Colunas Pivot


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)


L(B)={(1,1,0), (0,2,1), (0,0,0)}


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


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


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


5/10

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)


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 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)


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)


(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 v = 0


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)


(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 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


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


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

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Ad1-2014 Algebra Linear