METODO DE CRAMER ALGEBRA LINEAL

LAS ECUACIONES DADAS SON LAS SIGUIENTES:

𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝐷

𝑒𝑥 + 𝑓𝑦 + 𝑔𝑧 = 𝐻

𝑙𝑥 +𝑚𝑦 + 𝑛𝑧 = 𝑂

𝑑𝑜𝑛𝑑𝑒 𝑐𝑎𝑑𝑎 𝑢𝑛𝑎 𝑑𝑒 𝑙𝑎𝑠 𝑙𝑒𝑡𝑟𝑎𝑠 𝑒𝑥𝑐𝑒𝑝𝑡𝑜 𝑥, 𝑦 𝑦 𝑧 𝑠𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑠 (𝑛ú𝑚𝑒𝑟𝑜𝑠)

PARA ∆

∆=𝑎 𝑏 𝑐𝑒 𝑓 𝑔𝑙 𝑚 𝑛

=𝑎 𝑏 𝑐𝑒 𝑓 𝑔𝑙 𝑚 𝑛

𝑎 𝑏𝑒 𝑓𝑙 𝑚

= 𝑎 𝑓 𝑛 + 𝑏 𝑔 𝑙 + 𝑐 𝑒 𝑚 − [ 𝑙 𝑓 𝑐 + 𝑚 𝑔 𝑎 + (𝑛)(𝑒)(𝑏)

PARA ∆𝑥

∆𝑥 =𝐷 𝑏 𝑐𝐻 𝑓 𝑔𝑂 𝑚 𝑛

=𝐷 𝑏 𝑐𝐻 𝑓 𝑔𝑂 𝑚 𝑛

𝐷 𝑏𝐻 𝑓𝑂 𝑚

= 𝐷 𝑓 𝑛 + 𝑏 𝑔 𝑂 + 𝑐 𝐻 𝑚 − [ 𝑂 𝑓 𝑐 + 𝑚 𝑔 𝐷 +(𝑛)(𝐻)(𝑏)

PARA ∆𝑦

∆𝑦 =𝑎 𝐷 𝑐𝑒 𝐻 𝑔𝑙 𝑂 𝑛

=𝑎 𝐷 𝑐𝑒 𝐻 𝑔𝑙 𝑂 𝑛

𝑎 𝐷𝑒 𝐻𝑙 𝑂

= 𝑎 𝐻 𝑛 + 𝐷 𝑔 𝑙 + 𝑐 𝑒 𝑂 − [ 𝑙 𝐻 𝑐 + 𝑂 𝑔 𝑎 + (𝑛)(𝑒)(𝐷)

PARA ∆𝑧

∆𝑧 =𝑎 𝑏 𝐷𝑒 𝑓 𝐻𝑙 𝑚 𝑂

=𝑎 𝑏 𝐷𝑒 𝑓 𝐻𝑙 𝑚 𝑂

𝑎 𝑏𝑒 𝑓𝑙 𝑚

= 𝑎 𝑓 𝑂 + 𝑏 𝐻 𝑙 + 𝐷 𝑒 𝑚 − [ 𝑙 𝑓 𝐷 + 𝑚 𝐻 𝑎 +(𝑂)(𝑒)(𝑏)

PARA ENCONTRAR VALORES DE X, Y, Z

𝑥 =∆𝑥

∆

𝑦 =∆𝑦

∆

𝑧 =∆𝑧

∆

𝟐𝒙 − 𝟑𝒚 + 𝒛 = 𝟏𝟎𝒙 + 𝒚 − 𝒛 = 𝟐−𝒙 + 𝟐𝒚 − 𝟐𝒛 = 𝟑

∆=2 −3 11 1 −1−1 2 −2

=2 −3 11 1 −1−1 2 −2

2 −31 1−1 2

= 2 1 −2 + −3 −1 −1 + 1 1 2

− −1 1 1 + 2 −1 2 + −2 1 −3

= −4 − 3 + 2 − −1 − 4 + 6 = −5 − 1 = −5 − 1 = −6

∆𝑥 =10 −3 12 1 −13 2 −2

=10 −3 12 1 −13 2 −2

10 −32 13 2

= [ 10 1 −2 + −3 −1 3 + 1 2 2 ]

− 3 −1 1 + 2 −1 10 + −2 2 −3

= −20 + 9 + 4 − 3 − 20 + 12 = −7 − −5 = −7 + 5 = −2

∆𝑦 =2 10 11 2 −1−1 3 −2

=2 10 11 2 −1−1 3 −2

2 101 2−1 3

= 2 2 −2 + 10 −1 −1 + 1 1 3

− −1 2 −1 + 3 −1 2 + −2 1 10

= −8 + 10 + 3 − 2 − 6 − 20 = 5 − −28 = 5 + 28 = 33

∆𝑧 =2 −3 101 1 2−1 2 3

=2 −3 101 1 2−1 2 3

2 −31 1−1 2

= 2 1 3 + −3 2 −1 + 10 1 2

− −1 1 10 + 2 2 2 + 3 1 −3

= 6 + 6 + 20 − −10 + 8 − 9 = 32 − −11 = 32 + 11 = 43

𝑥 =∆𝑥

∆=

−2

−6=

1

3𝑦 =

∆𝑦

∆=

33

−6= −

33

6𝑧 =

∆𝑦

∆=

43

−6= −

43

6

∴ 𝑥 = 1 3 𝑦 = − 336 𝑧 = − 43

6

𝑿𝟏 − 𝟑𝑿𝟐 + 𝟒𝑿𝟑 = 𝟖𝟐𝑿𝟏 − 𝑿𝟐 − 𝟑𝑿𝟑 = 𝟒𝑿𝟏 + 𝑿𝟐 + 𝑿𝟑 = −𝟔

∆=1 −3 42 −1 −31 1 1

=1 −3 42 −1 −31 1 1

1 −32 −11 1

= 1 −1 1 + −3 −3 1 + 4 2 1

− 1 −1 4 + 1 −3 1 + 1 2 −3

= −1 + 9 + 8 − −4 − 3 − 6 = 16 − −13 = 16 + 13 = 29

∆𝑿𝟏 =8 −3 44 −1 −3−6 1 1

=8 −3 44 −1 −3−6 1 1

8 −34 −1−6 1

= 8 −1 1 + −3 −3 −6 + 4 4 −6

− −6 −1 4 + 1 −3 8 + 1 4 −3

= −8 − 54 + 16 − 24 − 24 − 12 = −46 − −12 = −46 + 12 = −34

∆𝑿𝟐 =1 8 42 4 −31 −6 1

=1 8 42 4 −31 −6 1

1 82 41 −6

= 1 4 1 + 8 −3 1 + 4 2 −6

− 1 4 4 + −6 −3 1 + 1 2 8

= 4 − 24 − 48 − 16 + 18 + 16 = −68 − 50 = −68 − 50 = −118

∆𝑿𝟑 =1 −3 82 −1 41 1 −6

=1 −3 82 −1 41 1 −6

1 −32 −11 1

= 1 −1 −6 + −3 4 1 + 8 2 1

− 1 −1 8 + 1 4 1 + −6 2 −3

= 6 − 12 + 16 − −8 + 4 + 36 = 10 − 32 = 10 − 32 = −22

𝑿𝟏 =∆𝑥∆= −34

29= − 34

29𝑿𝟐 =

∆𝑦∆= −118

29= − 118

29𝑿𝟑 = ∆𝑦

∆= −22

29= − 22

29

∴ 𝑿𝟏 = −3429 𝑿𝟐 = − 118

29 𝑿𝟑 = − 2229

BIBLIOGRAFIAS

Larson, Edwards, “INTRODUCCION AL ÁLGEBRA LINEAL”, 2006, Editorial LIMUSA, México, 752 Págs.