Post on 28-Jul-2015
METODO DE CRAMER ALGEBRA LINEAL
LAS ECUACIONES DADAS SON LAS SIGUIENTES:
ππ₯ + ππ¦ + ππ§ = π·
ππ₯ + ππ¦ + ππ§ = π»
ππ₯ +ππ¦ + ππ§ = π
πππππ ππππ π’ππ ππ πππ πππ‘πππ ππ₯ππππ‘π π₯, π¦ π¦ π§ π ππ ππππ π‘πππ‘ππ (πΓΊπππππ )
PARA β
β=π π ππ π ππ π π
=π π ππ π ππ π π
π ππ ππ π
= π π π + π π π + π π π β [ π π π + π π π + (π)(π)(π)
PARA βπ₯
βπ₯ =π· π ππ» π ππ π π
=π· π ππ» π ππ π π
π· ππ» ππ π
= π· π π + π π π + π π» π β [ π π π + π π π· +(π)(π»)(π)
PARA βπ¦
βπ¦ =π π· ππ π» ππ π π
=π π· ππ π» ππ π π
π π·π π»π π
= π π» π + π· π π + π π π β [ π π» π + π π π + (π)(π)(π·)
PARA βπ§
βπ§ =π π π·π π π»π π π
=π π π·π π π»π π π
π ππ ππ π
= π π π + π π» π + π· π π β [ π π π· + π π» π +(π)(π)(π)
PARA ENCONTRAR VALORES DE X, Y, Z
π₯ =βπ₯
β
π¦ =βπ¦
β
π§ =βπ§
β
ππ β ππ + π = πππ + π β π = πβπ + ππ β ππ = π
β=2 β3 11 1 β1β1 2 β2
=2 β3 11 1 β1β1 2 β2
2 β31 1β1 2
= 2 1 β2 + β3 β1 β1 + 1 1 2
β β1 1 1 + 2 β1 2 + β2 1 β3
= β4 β 3 + 2 β β1 β 4 + 6 = β5 β 1 = β5 β 1 = β6
βπ₯ =10 β3 12 1 β13 2 β2
=10 β3 12 1 β13 2 β2
10 β32 13 2
= [ 10 1 β2 + β3 β1 3 + 1 2 2 ]
β 3 β1 1 + 2 β1 10 + β2 2 β3
= β20 + 9 + 4 β 3 β 20 + 12 = β7 β β5 = β7 + 5 = β2
βπ¦ =2 10 11 2 β1β1 3 β2
=2 10 11 2 β1β1 3 β2
2 101 2β1 3
= 2 2 β2 + 10 β1 β1 + 1 1 3
β β1 2 β1 + 3 β1 2 + β2 1 10
= β8 + 10 + 3 β 2 β 6 β 20 = 5 β β28 = 5 + 28 = 33
βπ§ =2 β3 101 1 2β1 2 3
=2 β3 101 1 2β1 2 3
2 β31 1β1 2
= 2 1 3 + β3 2 β1 + 10 1 2
β β1 1 10 + 2 2 2 + 3 1 β3
= 6 + 6 + 20 β β10 + 8 β 9 = 32 β β11 = 32 + 11 = 43
π₯ =βπ₯
β=
β2
β6=
1
3π¦ =
βπ¦
β=
33
β6= β
33
6π§ =
βπ¦
β=
43
β6= β
43
6
β΄ π₯ = 1 3 π¦ = β 336 π§ = β 43
6
πΏπ β ππΏπ + ππΏπ = πππΏπ β πΏπ β ππΏπ = ππΏπ + πΏπ + πΏπ = βπ
β=1 β3 42 β1 β31 1 1
=1 β3 42 β1 β31 1 1
1 β32 β11 1
= 1 β1 1 + β3 β3 1 + 4 2 1
β 1 β1 4 + 1 β3 1 + 1 2 β3
= β1 + 9 + 8 β β4 β 3 β 6 = 16 β β13 = 16 + 13 = 29
βπΏπ =8 β3 44 β1 β3β6 1 1
=8 β3 44 β1 β3β6 1 1
8 β34 β1β6 1
= 8 β1 1 + β3 β3 β6 + 4 4 β6
β β6 β1 4 + 1 β3 8 + 1 4 β3
= β8 β 54 + 16 β 24 β 24 β 12 = β46 β β12 = β46 + 12 = β34
βπΏπ =1 8 42 4 β31 β6 1
=1 8 42 4 β31 β6 1
1 82 41 β6
= 1 4 1 + 8 β3 1 + 4 2 β6
β 1 4 4 + β6 β3 1 + 1 2 8
= 4 β 24 β 48 β 16 + 18 + 16 = β68 β 50 = β68 β 50 = β118
βπΏπ =1 β3 82 β1 41 1 β6
=1 β3 82 β1 41 1 β6
1 β32 β11 1
= 1 β1 β6 + β3 4 1 + 8 2 1
β 1 β1 8 + 1 4 1 + β6 2 β3
= 6 β 12 + 16 β β8 + 4 + 36 = 10 β 32 = 10 β 32 = β22
πΏπ =βπ₯β= β34
29= β 34
29πΏπ =
βπ¦β= β118
29= β 118
29πΏπ = βπ¦
β= β22
29= β 22
29
β΄ πΏπ = β3429 πΏπ = β 118
29 πΏπ = β 2229
BIBLIOGRAFIAS
Larson, Edwards, βINTRODUCCION AL ΓLGEBRA LINEALβ, 2006, Editorial LIMUSA, MΓ©xico, 752 PΓ‘gs.