Page 1

REPUBLICA BOLIVARIANA DE VENEZUELA MINSTERIO DEL PODER POPULAR PARA LA EDUCACION SUPERIOR “I.U.T.J.S. ANTONIO JOSE DE SUCRE BARQUISIMETO-LARA

Franklin Angulo 20929048 Informática 78 Ejercicio 1


 −6k − 15 4k − 18  x3  + −1 = 0 2k + 3   2k + 3 x3 = 0 −6k − 15 + 4k − 18 −1 = 0 2k + 3 −2k − 33 +1 = 0 2k + 3 −2k − 33 − 2k − 3 =0 2k + 3 −4k − 36 = 0 4k = −36 k = −9 u = [3, −9, −6], v = [−2,1, −6], w=[1,-7,4] x1u + x2 x = v x1 (3, −9, −6) + x2 (1, −7, 4) = (−2,1, −6) (3x1 , −9 x1 , −6 x1 ) + ( x2 , −7 x2 , 4 x2 ) = ( −2,1, −6) 3 x1 + x2 = −2

1

−9 x1 − 7 x2 = 1

2

−6 x1 + 4 x2 = −6

3

De 1 y 2 (3) 3x1 + x2 = −2 -9x 1 − 7 x2 = 1 9 x1 + 3 x2 = −6 − 9 x1 − 7 x2 = 1 (−1) -4x 2 =-5 4x 2 = 5 5 4 sustituyo x 2 en 1 x2 =

3 x1 + x2 = −2 5 = −2 4 5 3 x1 = −2 − 4 −13 3 x1 = 4 −13 x1 = 12 3 x1 +


Ejercicio 2

v1 = (2, −3, 0); v 2 = (−1, 0,3); v 3 = (0,3, −1) x1v1 + x2 v2 + x3v3 = 0 x1 (2,3, 0) + x2 (−1, 0,3) + x3 (0,3, −1) = 0 (2 x1 ,3 x1 , 0) + ( − x2 , 0,3 x2 ) + (0,3 x3 − x3 ) = (0, 0, 0) 2 x1 − x2 = 0

1

3 x1 + 3 x3 = 0

2

3 x2 − x3 = 0

3

De 3 3 x2 − x3 = 0 x3 = 3 x2 Sustituyo x 3 en 2 3 x1 + 3(3 x2 ) = 0 3 x1 + 9 x2 = 0

4

De 1 y 4 (9)

2x1 − x2 = 0 3x1 + 9 x2 = 0 18 x1 − 9 x 2 = 0 3x1 + 9 x 2 = 0 21x1 = 0

x1 = 0 Sustituyo x1 en 1 2 x1 − x2 = 0 210 > − x2 = 0 x2 = 0 Sustituyo x1 en 2 3x1 − 3 x3 = 0 310 > −3 x3 = 0 x3 = 0 Como x1 = x2 = x3 = 0 los vectores son L.I


Ejercicio 3.


x1 (2, −3, 0) + x2 (−1, 0,3) + x3 (0,3, −1) = ( a, b, c) (2 x1 , −3 x1 , 0) + (− x2 , 0,3 x2 ) + (0,3 x3 , − x3 ) = ( a, b, c) 2 x1 − x2 = a

1

−3x1 + 3 x3 = b

2

3 x2 − x3 = c

3

De 3 3 x2 − x3 = c x3 = 3 x2 − c

4

Sustituyo 4 en 2 −3x1 + 3(3x2 − c) = b −3x1 + 9 x2 − 3c = b −3x1 + 9 x2 = b + 3c

5

De 1 y 5 2x1 − x2 = a

(9)

-3x1 + 9 x2 = b + 3c 18 x1 − 9 x 2 = 9a -3x1 + 9 x 2 = b + 3c 15x1 = 9a + b + 3c 9a + b + 3c 15 Sustituyo x1 en 5 x1 =

-3x1 + 9 x2 = b + 3c  9a + b + 3c  −3   + 9 x2 = b + 3c 15  9a − b − 3c − z + 9 x2 = b + 3c 5 −9a − b − 3c + 45 x2 = b + 3c + 9a + b + 3c 45 x2 = 9a + 2b + 6c 9a + 2b + 6c 45 Sustituyo x1 en 2 x2 =

−3x1 + 3 x3 = b  9a + b + 3c  −3   + 3x3 = b 15 


−9a − b − 3c + 15 x3 =b 5 15 x3 = 5b + 9a + b + 3c 9a + 6b + 3c 15 Como son L.I y los genera los vectores dados forman una base en R 3 x3 =


Ejercicio 4

B1 = (u − 2v + w) B2 = (u + v ) B3 (u − v) x1 B1 + x2 B2 + x3 B3 = 0 x1 (u − 2v + w) + x2 (u + v) + x3 (u − v) = 0 u − 2v + w = 0

1

u+v =0

2

u −v = 0

3

De 2 y 3 u+v =0 u−v = 0 2u = 0 u=0 Sustituyo u en 2 m+v = 0 v=0 Sustituyo u y v en 1 u − 2v + w = 0 w=0 Como u=0 v=0 w=0 los vectores dados son C.I


algebra  
Read more
Read more
Similar to
Popular now
Just for you