Marices

Universidad Católica Andrés Bello
Escuela de Administración y Contaduría.
Cátedra Matemáticas IV
Profesor José Javier Salas
Nombre: Vanessa Maria Rodríguez Da Silva
Cedula: 19.819.332
Proyecto Matemáticas IV
Fecha de Entrega: 18 de Enero del 2011
Recibido por: ________________

Operaciones con Matrices
Sean las matrices
A=
n1 n2 n3 n4 n5 n6
n2 1 -1 1 1 n7
n3 0 -1 1 -1 n8
n4 -1 00 -1 n7
n3 0 -1 1 0 n6
n2 n1 n2 n3 n4 n5

D= C= B=
n1-n8 n1 n2 n3 n4 n1-n3
n2-n1 1 0 1 -1 n2-n4
n3-n2 1 n4 n5 1 n5-n7
n4-n3 0 n7 n6 0 n6-n8
N5-n6 1 1 -1 1 0
N7-n3 n8 n7 n6 n5 -1

Si sustituimos los valores, nos da:
A =

1 9 8 1 9 3
9 1 -1 1 1 3
8 0 -1 1 -1 2
1 -1 0 0 -1 3
8 0 -1 1 0 3
9 1 9 8 1 9 B =

-7
8
6
1
0
-1
C =

1 9 8 1
1 0 1-1
1 1 9 1
0 3 3 0
1 1 -1 1
2 3 3 9 D =

-1
8
-1
-7
6
1

Y el resultado es: _A=(D’x(((A2)’-3xA3)’xB)’
A2=
246 20 9 51 12 103
46 84 98 34 85 61
11 73 84 22 74 40
11 11 37 23 11 24
28 74 92 31 75 52
187 83 142 92 74 156

A3=
1572 2286 2854 1111 2277 1932
2849 525 650 801 427 1492
2314 190 217 562 106 1068
733 111 245 262 74 458
2529 347 451 685 255 1276
4158 1830 2601 17341688 2996

((A2)’-3*A3) =
-4470 -8527 -6933 -2148 -7575 -12371
-6812 -1491 -472 -299 -956 -5429
-8551 -1877 -567 -713 -1279 -7763
-3322 -2392 -1649 -763 -2044 -5178
-6803 -1207 -226 -191 -690 -5012
-5609 -4393 -3062 -1282 -3754 -8832

D’x(((A2)’-3*A3)’=
-64648 3585 10849 3382 7620 -25956

El resultado es _A=(D’x(((A2)’-3xA3)’xB)’=
575648
El resultado es =B’x((C x (2C’)) – 2A’)xD(C x (2C’))=

294 16 166 102 6 124
16 6 18 6 -2 -8
166 18 168 60 -12 82
102 6 60 36 0 36
6 -2 -12 0 8 22
124 -8 82 36 22 206

((C x (2C’)) – 2A’)=
292 -2 150 100 -10 106
-2 4 18 8 -2 -10
150 20 170 60 -10 64
100 4 58 36 -2 20
-12 -4 -10 2 8 20
118 -14 78 30 16 188

B’x((C x (2C’)) – 2A’)=
-1178 184 94 -270 -24 -606
El resultado es =B’x((C x (2C’)) – 2A’)xD=
3696

2.Determinantes
Demuestre que:
H=
1 0 1 0 1 0 1 0 1 0
1 1 0 1 0 1 0 1 0 1
-1 -1 0 0 -1 -1 0 0 -1 -1
0 1 1 1 -1 1 1 1 0 0
-1 -1 -1 1 -1 1 -1 1 0 1
1 1 1 1 1 1 1 1 1 1
1 2 -1 2 1 2 -1 2 0 0
1 -1 -1 1 -1 -1 1 -1 -1 1
0 -1 1 1 0 -1 0 0 -1 0
1 0 1 0 1 0 1 0 1 0

H=72m

Al tomar el método de la matriz de cofactores la última fila tenemos que:
h1=
0 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1
-1 00 -1 -1 0 0 -1 -1
1 1 1 -1 1 1 1 0 0
-1 -1 1 -1 1 -1 1 0 1
1 1 1 1 1 1 1 1 1
2 -1 2 1 2 -1 2 0 0
-1 -1 1 -1 -1 1 -1 -1 1
-1 1 1 0 -1 0 0 -1 0

det(h1)=0

h2=
1 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1
-1 0 0 -1 -1 0 0 -1 -1
0 1 1 -1 1 1 1 0 0
-1 -1 1 -1 1 -1 1 0 1
1 1 1 1 1 1 1 1 1
1 -1 2 1 2 -1 2 0 0
1 -1 1 -1 -1 1 -1 -1 1
0 1 1 0 -1 0 0 -1 0

det(h2)=-24

h3=
1 0 0 1 0 1 0 1 01 1 1 0 1 0 1 0 1
-1 -1 0 -1 -1 0 0 -1 -1
0 1 1 -1 1 1 1 0 0
-1 -1 1 -1 1 -1 1 0 1
1 1 1 1 1 1 1 1 1
1 2 2 1 2 -1 2 0 0
1 -1 1 -1 -1 1 -1 -1 1
0 -1 1 0 -1 0 0 -1 0

det(h3)=0

h4=
1 0 1 1 0 1 0 1 0
1 1 0 0 1 0 1 0 1
-1 -1 0 -1 -1 0 0 -1 -1
0 1 1 -1 1 1 1 0 0
-1 -1 -1 -1 1 -1 1 0 1
1 1 1 1 1 1 1 1 1
1 2 -1 1 2 -1 2 0 0
1 -1 -1 -1 -1 1 -1 -1 1
0 -1 1 0 -1 0 0 -1 0
det(h4)=-24h5=
1 0 1 0 0 1 0 1 0
1 1 0 1 1 0 1 0 1
-1 -1 0 0 -1 0 0 -1 -1
0 1 1 1 1 1 1 0 0
-1 -1 -1 1 1 -1 1 0 1
1 1 1 1 1 1 1 1 1
1 2 -1 2 2 -1 2 0 0
1 -1 -1 1 -1 1 -1 -1 1
0 -1 1 1 -1 0 0 -1 0
det(5)=-24

h6=
1 0 1 0 1 1 0 1 0
1 1 0 1 0 0 1 0 1
-1 -1 0 0 -1 0 0 -1 -1
0 1 1 1 -1 1 1 0 0
-1 -1 -1 1 -1 -1 1 0 1
1 1 1 1 1 1 1 1 1
1 2 -1 2 1 -1 2 0 0
1 -1 -1 1 -1 1 -1 -1 1
0 -1 1 1 0 0 0-1 0
det(h6)=48

h7=
1 0 1 0 1 0 0 1 0
1 1 0 1 0 1 1 0 1
-1 -1 0 0 -1 -1 0 -1 -1
0 1 1 1 -1 1 1 0 0
-1 -1 -1 1 -1 1 1 0 1
1 1 1 1 1 1 1 1 1
1 2 -1 2 1 2 2 0 0
1 -1 -1 1 -1 -1 -1 -1 1
0 -1 1 1 0 -1 0 -1 0
det(h7)=-24

h8=
1 0 1 0 1 0 1 1 0
1 1 0 1 0 1 0 0 1
-1 -1 0 0 -1 -1 0 -1 -1
0 1 1 1 -1 1 1 0 0
-1 -1 -1 1 -1 1 -1 0 1
1 1 1 1 1 1 1 1 1
1 2 -1 2 1 2 -1 0 0
1 -1 -1 1 -1...