TALLER #3 ALGEBRA LINEAL
Enviado por camilo andres suarez sanchez • 26 de Abril de 2020 • Exámen • 3.587 Palabras (15 Páginas) • 171 Visitas
[pic 1]
TALLER #3 ALGEBRA LINEAL
CAMILO ANDRÉS SUAREZ SÁNCHEZ ID: 725358[pic 2]
DAVID ENRIQUE LEÓN GUALTEROS ID: 728357
CORPORACION UNIVERSITARIA MINUTO DE DIOS FACULTAD DE INGENIERIA CIVIL
ALGEBRA LINEAL NRC: 10136
GIRARDOT-CUNDINAMARCA 2020-04
[pic 3]
TALLER #3 ALGEBRA LINEAL
CAMILO ANDRÉS SUAREZ SÁNCHEZ ID: 725358
DAVID ENRIQUE LEÓN GUALTEROS ID: 728357[pic 4]
DIEGO MAURICIO PARRA LAGUNA DOCENTE
CORPORACION UNIVERSITARIA MINUTO DE DIOS FACULTAD DE INGENIERIA CIVIL
ALGEBRA LINEAL NRC: 10136
GIRARDOT-CUNDINAMARCA 2020-04
[pic 5]
- Dadas las siguientes matrices, resolver las operaciones descritas a continuación
5 1 −9
1 7 10
−3 9 6
A = [2
6 1 ]
B = [4
6 −5]
C = [−2 5 7]
7 −1 −3 3𝑋3
8 12 8
3𝑋3
−1 1
8 3𝑋3
2 3 2 −8
1 3 0
D = [9 0 5 3 ]
1 −5 6 0 3𝑋4
E = [4 −7 2]
2𝑋4
F = [1 4 0 −6]1𝑋4
−1 0 1 7 3 7[pic 6]
G = [ 4[pic 7]
8
2 4 6 ]
3 −1 5 3𝑋4
5 −1
= [ ]
8 2
2𝑋2
I = [−5
6
2
8 ]
4
−1 4𝑋2
J = [ 1 −9 6]
−5 7 3
2𝑋3
[pic 8]
Resolver
5 1 −9
1 7 10
−3 9 6
1. 1⁄3 A + 3⁄5 B – C A = [2
6 1 ]
B = [4
6 −5]
C = [−2 5 7]
7 −1 −3 3𝑋3
8 12 8
3𝑋3
−1 1
8 3𝑋3
5⁄ 1⁄ −3
3⁄5
21⁄5 6
3 3
- 1⁄3 A = 2⁄3[pic 9]
2 1⁄3
+ 3⁄5 B =
12⁄5
18⁄5 −3
7⁄ −1⁄ −1
24⁄ 36⁄ 24⁄
[ 3 3 ]
[ 5 5 5]
34⁄15[pic 10]
68⁄15 3
−3 9 6
A*B =
46⁄15
28⁄5 −8⁄3
- C = [−2 5 7]
[107⁄ 36⁄ 19⁄ ]
−1 1 8 3𝑋3
15 5
5 3𝑋3
79⁄15 −67⁄15 −3
AB – C =
76⁄15
3⁄5
−29⁄3
RTA
122⁄ 88⁄ −21⁄
[ 15 15
5]3𝑋3
1 7 10
−3 9 6
5 1 −9
2. 2B – 3C + 1⁄2 A B = [4
6 −5]
C = [−2 5 7]
A = [2
6 1 ]
8 12 8
3𝑋3
−1 1
8 3𝑋3
7 −1 −3 3𝑋3
2 14 20 −9 27 18
...