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.
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