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# Algebra Lineal

Summary: Solucionario de ejercicios de algebra lineal

Ejercicios 4.2

n-vectores.

1. Determine u+v,u-v,2u,3u-2vu+v,u-v,2u,3u-2v si

a) u = ( 1 , 2 , - 3 ) , v = ( 0 , 1 , - 2 ) u = ( 1 , 2 , - 3 ) , v = ( 0 , 1 , - 2 )

b)

u = ( 4 , - 2 , 1 , 3 ) , v = ( - 1 , 2 , 5 , - 4 ) u = ( 4 , - 2 , 1 , 3 ) , v = ( - 1 , 2 , 5 , - 4 )
(1)

Soluci'on

a)

u = vector([1,2,-3])

v = vector([0,1,-2])

redsage] blueu+v

1 , 3 , - 5 1 , 3 , - 5

redsage] blueu-v

1 , 1 , - 1 1 , 1 , - 1

redsage] blue2*u

2 , 4 , - 6 2 , 4 , - 6

redsage] blue3*u-2*v

3 , 4 , - 5 3 , 4 , - 5

b)

redsage] blueu = vector([4,-2,1,3])

redsage] bluev = vector([-1,2,5,-4])

redsage] blueu+v

3 , 0 , 6 , - 1 3 , 0 , 6 , - 1

redsage] blueu-v

5 , - 4 , - 4 , 7 5 , - 4 , - 4 , 7

redsage] blue2*u

8 , - 4 , 2 , 6 8 , - 4 , 2 , 6

redsage] blue3*u-2*v

14 , - 10 , - 7 , 17 14 , - 10 , - 7 , 17

3. Sean

u = ( 1 , - 2 , 3 ) , v = ( - 3 , - 1 , 3 ) , w = ( a , - 1 , b ) , x = ( 3 , c , 2 ) u = ( 1 , - 2 , 3 ) , v = ( - 3 , - 1 , 3 ) , w = ( a , - 1 , b ) , x = ( 3 , c , 2 )
(2)

determine a, b y c de modo que

a)

w = 1 2 u w = 1 2 u
(3)

b)

w + v = u w + v = u
(4)

c)

w - u = v w - u = v
(5)

Solucion.

Sabiendo las propiedades de la suma, resta y multiplicacion de matrices sabemo que para

a)

w = 1 2 u w = 1 2 u
(6)
( a , - 1 , b ) = 1 2 ( 1 , - 2 , 3 ) ( a , - 1 , b ) = 1 2 ( 1 , - 2 , 3 )
(7)
( a , - 1 , b ) = ( 1 2 ( 1 ) , 1 2 ( - 2 ) , 1 2 ( 3 ) ) ( a , - 1 , b ) = ( 1 2 ( 1 ) , 1 2 ( - 2 ) , 1 2 ( 3 ) )
(8)
( a , - 1 , b ) = ( 1 2 , - 1 , 3 2 ) ( a , - 1 , b ) = ( 1 2 , - 1 , 3 2 )
(9)

por lo tanto, podemos decir que

a = 1 2 y b = 3 2 a = 1 2 y b = 3 2
(10)

b)

w + v = u w + v = u
(11)
( a , - 1 , b ) + ( - 3 , - 1 , 3 ) = ( 1 , - 2 , 3 ) ( a , - 1 , b ) + ( - 3 , - 1 , 3 ) = ( 1 , - 2 , 3 )
(12)
( a + ( - 3 ) , - 1 + ( - 1 ) , b + 3 ) = ( 1 , - 2 , 3 ) ( a + ( - 3 ) , - 1 + ( - 1 ) , b + 3 ) = ( 1 , - 2 , 3 )
(13)

eso es

a - 3 = 1 a - 3 = 1
(14)
b + 3 = 3 b + 3 = 3
(15)

por lo tanto

a = 4 a = 4
(16)
b = 0 b = 0
(17)

c)

w + x = v w + x = v
(18)
( a , - 1 , b ) + ( 3 , c , 2 ) = ( - 3 , - 1 , 3 ) ( a , - 1 , b ) + ( 3 , c , 2 ) = ( - 3 , - 1 , 3 )
(19)
( a + 3 , - 1 + c , b + 2 ) = ( - 3 , - 1 , 3 ) ( a + 3 , - 1 + c , b + 2 ) = ( - 3 , - 1 , 3 )
(20)

eso es

a + 3 = - 3 a + 3 = - 3
(21)
c - 1 = - 1 c - 1 = - 1
(22)
b + 2 = 3 b + 2 = 3
(23)

por lo tanto

a = - 6 a = - 6
(24)
c = 0 c = 0
(25)
b = 1 b = 1
(26)

5. Sean

u = ( 4 , 5 , - 2 , 3 ) , v = ( 3 , - 2 , - 4 , 1 ) , x = ( - 3 , 2 - 5 , 3 ) , c = 2 y d = 3 u = ( 4 , 5 , - 2 , 3 ) , v = ( 3 , - 2 , - 4 , 1 ) , x = ( - 3 , 2 - 5 , 3 ) , c = 2 y d = 3
(27)

Verifique las propiedades (a) a (h) del teorema 4.2

Solucion

a)

u + v = v + u u + v = v + u
(28)

redsage] blueu=vector([4,5,-2,3])

redsage] bluev=vector([3,-2,-4,1])

redsage] blueu+v

7 , 3 , - 6 , 4 7 , 3 , - 6 , 4

redsage] bluev+u

7 , 3 , - 6 , 4 7 , 3 , - 6 , 4

redsage] blueu+v==v+u

1

b)

u + ( v + x ) = ( u + v ) + x u + ( v + x ) = ( u + v ) + x
(29)

redsage] blueu = vector([4,5,-2,3])

redsage] bluev = vector([3,-2,-4,1])

redsage] bluex = vector([-3,2,-5,3])

redsage] blueu+(v+x)

4 , 5 , - 11 , 7 4 , 5 , - 11 , 7

redsage] blue(u+v)+x

4 , 5 , - 11 , 7 4 , 5 , - 11 , 7

c)

Existe un vector 0 en R n tal que u + 0 = 0 + u = u para toda u en R n . Existe un vector 0 en R n tal que u + 0 = 0 + u = u para toda u en R n .
(30)

redsage] blueu = vector([4,5,-2,3])

redsage] bluea=0

redsage] blueu+0

4 , 5 , - 2 , 3 4 , 5 , - 2 , 3

redsage] blue0+u

4 , 5 , - 2 , 3 4 , 5 , - 2 , 3

redsage] blueu

4 , 5 , - 2 , 3 4 , 5 , - 2 , 3

redsage] blueu+0==0+u

1

d)

para cada vector u en R n , existe un vector - u en R n tal que u + ( - u ) = 0 . para cada vector u en R n , existe un vector - u en R n tal que u + ( - u ) = 0 .
(31)

redsage] blueu = vector([4,5,-2,3])

redsage] blueu

4 , 5 , - 2 , 3 4 , 5 , - 2 , 3

redsage] blue-u

- 4 , - 5 , 2 , - 3 - 4 , - 5 , 2 , - 3

redsage] blueu+(-u)

0 , 0 , 0 , 0 0 , 0 , 0 , 0

redsage] blueu+(-u)==0

1

e)

c ( u + v ) = c u + v u c ( u + v ) = c u + v u
(32)

redsage] blueu = vector([4,5,-2,3])

red bluev = vector([3,-2,-4,1])

red bluec=2

red bluec*(u+v)

14 , 6 , - 12 , 8 14 , 6 , - 12 , 8

red bluec*u+c*v

14 , 6 , - 12 , 8 14 , 6 , - 12 , 8

red bluec*(u+v)==c*u+c*v

1

f)

( c + d ) u = cu + du ( c + d ) u = cu + du
(33)

red blueu = vector([4,5,-2,3])

red bluec=2

red blued=3

red blue(c+d)*u

20 , 25 , - 10 , 15 20 , 25 , - 10 , 15

red bluec*u+d*u

20 , 25 , - 10 , 15 20 , 25 , - 10 , 15

red blue(c+d)*u==c*u+d*u

1

g)

c ( du ) = ( cd ) u c ( du ) = ( cd ) u
(34)

red blueu = vector([4,5,-2,3])

red bluec=2

red blued=3

red bluec*(d*u)

24 , 30 , - 12 , 18 24 , 30 , - 12 , 18

red blue(c*d)*u

24 , 30 , - 12 , 18 24 , 30 , - 12 , 18

red bluec*(d*u)==(c*d)*u

1

h)

1 u = u 1 u = u
(35)

red blueu = vector([4,5,-2,3])

red blue1*u

4 , 5 , - 2 , 3 4 , 5 , - 2 , 3

red blue1*u==u

1

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