algebra todos ejercicios

7/24/2019 Algebra Todos Ejercicios

1/21

1.- Si se tiene el conjunto de vectores {[2,-1,3]; [-4,2,-4]; [8,0,1]} Determine

usted ue es line!lmente inde"endiente en #3$ com"rue%e sus result!dos.

c=

[

2 4 81 2 0

3 4 1

0

0

0

]

1

2R

1c=

[

1 2 00 0 1

3 4 0

0

0

0

]R

3-3#1

c=[ 1 2 41 2 0

3 4 1

0

0

0]R 2 c=[

1 2 00 0 1

0 2 0

0

0

0]R 1

c=[1 2 40 0 4

3 4 1

0

0

0]R 1-#2 c=[

1 0 0

0 0 1

0 2 0

0

0

0]12 R 3

c=[1 2 00 0 4

3 4 1

0

0

0]14 R 2 c=[

1 0 0

0 0 1

0 1 0

0

0

0]R2 R3

c=[

1 2 00 0 1

3 4 1

0

0

0]R 3-#2 c=[

1 0 0

0 1 0

0 0 1

0

0

0]

'1(0

'2(0

'3(0

'om"ro%!ci)n*

'1 [ 2

13] &'2 [

42

4] &'3 [8

0

1] ( [

0

0

0]

0[ 21

3] &0 [4

2

4] &0 [80

1] ( [00

0] [

0

0

0]=[

0

0

0]


2/21

+or lo t!nto el S... es line!lmente inde"endiente.


3/21

2.- Si se tiene el sistem!*

2#1&4#2-2#3&4#4(0#1&14#2&2#3&8#4(0Determine usted sus vectores $ demuestre ue son line!lmente

inde"endientes.

[2 4 26 14 2 4 08 0][ 12 ] #1

[1 2 16 14 2 2 08 0] [ 12 ] #2

[1 2 13 7 1 2 04 0] #2-3#1

[1 2 10 1 4 2 02 0] #1

[1 3 30 1 4 0 02 0]#1&3#2&3#3(0#2&4#3-2#4(0

Des"ej!mos ! #3

3#3( - #1 / 3#24#3( - #2& 2#4D!mos v!lores ! #3(0

-#1-3#2(0-#2&2#4(0D!mos v!lores ! #2(0 "or lo t!nto #4$ #1sern iu!l ! 0. ntonces

tenemos*

#1(0#2(0#3(0#4(0'om"ro%!ndo en el S... oriin!l tenemos*

20&40-20&40(0


4/21

0&140&20&80(0

+or lo t!nto son inde"endientes.


5/21

3.-Si se tienen los vectores* 1( [ 2

60] 2( [

6

0

8] 3( [

22

1224

]; 'onstru$!usted el S... 5omo6neo $ demuestre ue el conjunto de vectores es

line!lmente de"endiente en #3

$ com"rue%e sus result!dos.

'1 [ 2

60] &'2 [

6

0

8] &'3 [

22

12

24] ( [

0

0

0]

'( [ 2 6 22

6 0 120 8 24

0

0

0] r2&3r1 '( [

2 6 22

0 2 6

0 0 0

0

0

0] r1-3r2

'( [ 2 6 22

6 18 540 8 24

0

0

0] r2( 19 ) '( [

2 0 4

0 2 6

0 0 0

0

0

0] r1( 12 )

'( [2 6 22

0 2 6

0 8 24

0

0

0] r3-4r2 '( [

1 0 2

0 2 6

0 0 0

0

0

0] r2( 12 )

'( [1 0 20 1 3

0 0 0

00

0] r2(1

2 ) 7!tri reducid! "or el m6todode 9!uss :ord!n

'1&0 '2&2'3(0 '1(-2'30'1&1'2&3'3(0 '2-3'3'3(1

'1(-2

'2(-3

'1 [ 2

60] &'2 [

6

0

8] &'3 [

22

12

24] ( [

0

0

0]


6/21

-2 [ 2

60] -3 [

6

0

8] &1 [

22

12

24] ( [

0

0

0]

4 18 +2212 0 120 24 +24

000

+or lo t!nto el siuiente S... es de"endiente en #3.

4.- si se tiene el S...

1&22-3&24(0


7/21


8/21

?.-si se tiene el conjunto de vectores ,-4,4,,-2,8,2,0,. Determine ue es

line!lmente inde"endiente en #. com"rue%e sus result!dos

2

@( -4 A( -2 '( 0

4 8

SOLUCION:

6 6 2 0 r2= -4 -2 0 0

-4 -2 0 0 2/3r1= 4 4 4/3 0

4 8 6 0 0 2 4/3 0

'om"ro%!ci)n'omo el sistem! tiene soluci)n Bnic!c1( 0, c2( 0 $ c3( 0Se deduce ue l! Bnic! Corm! de com%in!r los vectores Es "!r! ue den elvector cero es l! ue tiene todos los coeFcientes cero. +or t!nto, el conjunto devectores es line!lmente inde"endiente.


9/21

.- Si se tiene el conjunto de vectores 1,2,?,1,-2,1,2,1,4. Determine ue es

line!lmente inde"endiente en #3, com"rue%e sus result!dos.

Comprobacin'omo el sistem! tiene soluci)n Bnic!c1( 0, c2( 0 $ c3( 0Se deduce ue l! Bnic! Corm! de com%in!r los vectores Es "!r! ue den elvector cero es l! ue tiene todos los coeFcientes cero. +or t!nto, el conjunto devectores es line!lmente inde"endiente.


10/21


11/21


12/21


13/21

10.-Si se tienen los vectores* u12,-1,0; u2(4,-1,0; u3(4,0,-1. ncuentre ud.

un! %!se ortonorm!l del es"!cio vectori!l en #3medi!nte el m6todo de

9r!m- Sc5midt $ com"rue%e sus result!dos.

1( [2,-1,0]

2( [4,-1,0] -[2,1,0 ] [4,1,0][2,1,0 ] [2,1,0] [2,-1,0]

2( [4,-1,0] /[ 8+1+0 ][ 4+1+0 ] [2,-1,0]

2( [4,-1,0] /[9][5] [2,-1,0]

2( [4,-1,0] / [ 185 ;95 ; 0]

2([ 2

5 ,

4

5 ,0

3( [4,0,1] -

[25 ,45 ,0][4,0,1]

[2

5,4

5,0 ][

2

5,

4

5,0] [

2

5,

4

5,0

] -[ 2,1,0 ][4,0,1]

[2,1,0 ] [2,1,0] [2,-1,0]

3( [4,0,1] -[85+0+0]

[ 425 + 1625+0] [ 25 , 45 ,0] - [8+0+0 ][ 4+1+0 ] [2,-1,0]

3( [4,0,1] - [8

5 ][45 ] [ 25 , 45 , 0] - [8 ][5 ] [2,-1,0] ( [4,0,1] -2

[ 25 , 45 ,0] - [8 ][5 ] [2,-1,0]


14/21

3( [4,0,1] - [ 45 , 85 ,0] - [ 165 ,85 ,0 ] ( [ 165 ,85 ,1] - [ 165 ,85 ,0 ]

3( [0,0,1]

1( [2,-1,0] 2([ 2

5 ,

4

5 ,0 3( [0,0,1]

GG1GG( (2)2+(1)2+(0)2 ( 4+1+0 ( 5

91([

2

5;

8

5;

0

5]

GG2GG( (2

5)

2

+(4

5)

2

+(0)2 ( 425 + 1625+0 ( 45

92(

[

2

5

4

5

;

4

5

4

5

; 0

4

5

]

GG3GG( (0)2+(0)2+(1)2 ( 0+0+1 ( 1

93([

0

1;

0

1;

1

1]

'om"ro%!ci)n*

91H92( [

2

5;

8

5;

0

5][

2

5

45;

4

5

45;

0

45] ( [

2

5

2

5+0] ( 0


15/21

91H93( [

2

5;

8

5;

0

5][

0

1;

0

1;

1

1] ([0&0&0] ( 0

92H93( [

2

5

45;

4

5

45;

0

45][

0

1 ; 0

1 ; 1

1 ] ( [0&0&0] ( 0


16/21


17/21

13.- Si se tienen l!s %!ses I1( 2,2,3,1 ,I2 (2,0,1,1) ,I3 (3,2,0,1)

medi!nte t6cnic! de 9r!m- Sc5midt constru$! usted un! %!se ortonorm!l del

es"!cio vectori!l v en r4 $ com"rue%e sus result!dos.

NORMALIZAR V1

U1=

V1

|V1|=

(2,2,3,1)

22+22+32+12=(2,2,3,1)

32=( 232 ,

2

32,

3

32,1

32 )

OBTENER EL VECTOR ORTOGONAL A V2

U2=V2(V2 .U1)U1

(2,0,1,1 )[ (2,0,1,1 ) .( 232 , 232 , 332 , 132 )][ 232 , 232 , 332 , 132 ]

( 2,0,1,1)[4 2323 ,0,32323 ,2323 ][ 232 , 2

32,

3

32,132 ]

(2,0,1,1 ) [0 ] [ 2

32 , 2

32 , 3

32 ,132 ]

= (2,0,1,1)

NORMALIZAR V2

U1=

V2

|V2|=

(2,0,1,1 )

6=

( 26 , 0,1

6,

1

6 )

OBTENER EL VECTOR ORTOGONAL A V1 Y V2


18/21

U3=V3 (V3. UI)U1(V3.U2 )U2

=

(3,2,0,1)

[(3,2,0,1) .

[ 2

32,

2

32,

3

32,1

32 ]] [ 2

32,

2

32,

3

32,1

32 ]

[( 3,2,0,1) .

( 2

6,0,

1

6,

1

6 )]

(3,2,0,1)[2 , 223 ,0,26 ][ 232 , 2

32,

3

32,132 ](6 ,0,0

6

6)( 26 ,0,16

, 1

6 )

(3,2,0,1)[(322)( 232 , 232 , 332 , 132 )][(766 )( 26 ,0,16 , 16 )]

( 3,2,0,1 )+[34623 ,34623 ,94646 , 34646 ]+[73 ,0, 76 ,76 ]

(3,2,0,1 )+(3.21789,4623 ,.1603,0.7243)

=(-.21798,1.115348,-0.1603,0.2757)

NORMALIZAR V2

U1=

V3

|V3|=(.21798,1.115348,0.1603,0.2757)

1.1803

=-0.1846,0.9424,-.1358,0.2335

COMPROBACON!


19/21

V1.V2=0

( 232 , 2

32,

3

32,132 ) .(

2

6,0,

16

, 1

6 )=[23

9

3

6

3

18 ]=239 239 =0

V2.V3=0

( 26 ,0,1

6,

1

6 ). (0.1846,0 .9424,.1358,0.2335 )=(0.1507+0.0554+0.0953 )=0.1507+ .1507=0

14.- .- Si se tienen l!s %!ses I1( 4,4,6,2 ,I2 (4,0,2,2) ,I3 (6,4,0,2)

medi!nte t6cnic! de 9r!m- Sc5midt constru$! usted un! %!se ortonorm!l del

es"!cio vectori!l v en r4 $ com"rue%e sus result!dos.

NORMALIZAR V1

U1=

4,4,6,2V1

|V1|=

(4,4,6,2)

42+42+62+22=

62=( 23 , 23 ,22 ,26 )

OBTENER EL VECTOR ORTOGONAL A V2

U2=V2(V2 .U1)U1

(4,0,2,2 )

[( 4,0,2,2 ) .

(2

3

,2

3

,2

2

,2

6

)](2

3

,2

3

,2

2

,2

6

) ( 4,0,2,2 )[4 23 ,0,222 ,23 ]( 23 , 23 ,22 ,26 )


20/21

( 4,0,2,2 )[ 0 ] [ 232 , 2

32,

3

32,132 ]

= (4,0,2,2 )

NORMALIZAR V2

U1=

V2

|V2|=

(4,0,2,2 )

26=

( 63 , 0,66 ,66)

OBTENER EL VECTOR ORTOGONAL A V1 Y V2

U3=V3 (V3.UI)U1(V3.U2 )U2

=

(6,4,0,2 )[(6,4,0,2 ) .( 23 , 23 ,22 ,26 )][( 23 ,23 , 22 ,26 )][ ( 6,4,0,2 ) .( 63 ,0,66 ,66)](63 ,

(6,4,0,2 )[[22 ,

4 23 ,0,

23

](23 ,

23 ,

22 ,

26

)]

[(26 ,0,0,

63

)(63 ,0,

66 ,

66 )

] (6,4,0,2)[(32 )( 23 ,23 , 22 ,26 )][( 763 )( 63 ,0,66 , 66)]

( 6,4,0,2)+ [2,2,3,1 ]+[143 ,0, 73 ,73 ]

(6,4,0,2 )+(203

,2,23,4

3 )

=(23 ,2,23 , 23 )


21/21

NORMALIZAR V3

U3= V3|V3|=(

2

3

,2,2

3

,2

3

)433

=(3

6,32

,36,36)

COMPROBACON!

V1.V2=0

( 23 , 23 ,22 ,26 ).( 63 ,0, 66 , 66)=[2 39 36 318]=0

V2.V3=0

( 63 , 0,66 ,66) .(36 , 32 ,36 ,36)=[26 , 212 ,212 ]=26 + 26 =0

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