Matrices

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  • Publicado : 23 de mayo de 2010
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DESARROLLO ACTIVIDAD No.4
1. Dadas las matrices:

A=23-7-154404; B=-1045/3125-19 y C=7-4-4511/2866

a. A+B+C

23-7-154404+-1045/3125-19+7-4-4511/2866=8-1-717/3713/217519b. A – 3B – 2C

23-7-154404-3-1045/3125-19-27-4-4511/2866=23-7-154404+30-12-5-3-6-153-27+-1488-10-2-1-16-12-12=-911-11-160-3-27-9-35

2. Dadas las matrices

A=-154205-36;B=5-4-23;C=-926 y D= 1050812
c. AB
-1542 05-36.5-4-23= 1036

C11= [-1 5 0 5].[5 -4 -2 3] = (-5 -20 + 0 + 15) = 10
C21= [4 2 -3 6].[5 -4 -2 3] = (20 -8 + 6 + 18) = 36

d. BC
5-4-23.-926=-4536 18-2710-8 -46 30-24-1218

C11= [5].[-9] = -45
C12 = [5].[2] = 10
C13 = [5].[6] = 30

C21= [-4].[-9] = 36
C22 = [-4].[2] = -8
C23 = [-4].[6] = -24

C31= [-2].[-9] = 18
C32 =[-2].[2] = -4
C33 = [-2].[6] = -12

C41= [3].[-9] = -27
C42 = [3].[2] = 6
C43 = [3].[6] = 18

e. CD
-926.1050812= 84-17

C11= [-9 2 6].[10 0 1] = (-90 + 0 + 6) = 84
C12= [-9 2 6].[5 82] = (-45 + 16 +12) = -17

f. DA

1050812.-1542 05-36= 10327 60169 -15-24-6 804817

C11= [10 5].[-1 4] = (-10+20) = 10
C12= [10 5].[5 2] = (50 + 10) = 60
C13= [10 5].[0 -3]= (0 – 15) = -15
C14= [10 5].[5 6] = (50 + 30) = 80

C21= [0 8].[-1 4] = (0+32) = 32
C22= [0 8].[5 2] = (0 + 16) = 16
C23= [0 8].[0 -3] = (0 – 24) = -24
C24= [0 8].[5 6] = (0 + 48) = 48C31= [1 2].[-1 4] = (-1+8) = 7
C32= [1 2].[5 2] = (5 + 4) = 9
C33= [1 2].[0 -3] = (0 – 6) = -6
C34= [1 2].[5 6] = (5 + 12) = 17

3. De las matrices del punto 1. hallar:

g. 4A423-7-154404=812-28-4201616016

h. -5B

-5-1045/3125-19=50-20-25/3-5-10-255-45

i. 8C

87-4-4511/2866=56-32-324084644848

4. Solucionar el sistema de tres ecuacionessimultáneas con tres incógnitas por medio de matrices, empleando la regla de Kramer y Sarrus.

1. X + Y + Z = 4
2. 2X -3Y + 5 Z = -5
3. 3X +4Y +7Z = 10
Δ=1112-3531241-3715
α=-21+8+15= 2...
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