Sistemas Consistentes e Inconsistentes PDF

Title Sistemas Consistentes e Inconsistentes
Author Martin Avalos
Course Álgebra Lineal
Institution Instituto Tecnológico de Tijuana
Pages 3
File Size 101.9 KB
File Type PDF
Total Downloads 27
Total Views 138

Summary

Ejercicios de Sistemas Consistentes e Inconsistentes...


Description

−2x1 + x3 = 0 x 2 + 3x 3 = 1 x1 − x2 = −3  −2 0 1 R= 0 1 3 1 −1 0

  1 R1 → R1 /2 |0 ≈ 0 → |1  → |−3 1

1 R= 0 1

0 1 −1

− 12 3 0

  |0 1 → ≈ 0 → |1  0 R 3 → −R 3 + R 1 |−3



0 1 1

− 12 3 − 12

  → |0 1 → |1  ≈ 0 R 3 → −R 2 + R 3 |3 0



0 1 0

− 12 3 − 72

  1 → |0 ≈ 0 |1  → R3 → − 72 ∗ R3 |2 0



1 R= 0 0 1 R= 0 0 

1 R= 0 0 

1 R= 0 0

−21 3 1

0 1 0 0 1 0

−21 3 1

− 21 3 0

0 1 −1

− 12 3 − 12

0 1 1

− 12 3 − 72

0 1 0 0 1 0

 |0 |1  |−3

− 12 3 1

 |0 |1  |2  |0   |1 − 4 7

  1 → |0  R2 → −3R3 + R2 ≈  0  |1 − 4 0 → 7

0 1 0

− 12 0 1

  |0 1 R1 → 12 R3 + R1  ≈ 0 →  |1 − 4 0 → 7

0 1 0

0 0 1

2 4 −2(− ) + (− ) = 0 7 7

(

(

19 4 ) + 3(− ) = 1 7 7

19 −2 )−( ) = −3 7 7

 |0 |1  |3

 |0  19   7 − 4 7  −2   7  19    7 − 4 7

x1 + x2 − x3 = 7 4x1 − x2 + 5x3 = 4 6x1 + x2 + 3x3 = 20    → 1 1 1 −1 |7 R =  4 −1 5 |4  R2 → −4R1 + R2 ≈  0 6 6 1 3 |20 → 

1 R= 0 6

1 −5 1



1 R= 0 0 

1 R= 0 0

1 −5 −5 1 1 −5

−1 9 3

1 −5 1

  |7 → 1 ≈ 0 |−24  → |20 R3 → −6R1 + R3 0

  1 → −1 |7 9 |−24  R2 → R2 / − 5 ≈  0 → −9 |−22 0 −1 − 59 −9

1 −5 −5 1 1 −5

  7 → 1 |24    0 ≈ → 5 R3 → 5R1 + R3 |−22 0

1 1 0

−1 9 3

 |7 |−24  |20

 −1 |7 9 |−24  −9 |−22 −1 − 95 −9 −1 − 95 0

 7 |24   5 |−22  7 |24   5 |2

0x3 = 2

x 1 + 2x 2 − 4x 3 = 4 −2x1 − 4x2 + 8x3 = −8    1 2 −4 |4 1 2 → R= ≈ R2 → 2R1 + R2 0 0 −2 −4 8 |−8 x1 = 4 − 2x2 + 4x3 → x1 = 4 − 2s + 4t x2 = s x3 = t        −2 x1 4 4  x2  =  0  + s  1  + t  0  x3 1 0 0 

−4 |4 0 |0



2x1 + 6x2 − 4x3 + 2x4 = 4 x1 − x3 + x4 = 5 −3x1 + 2x2 − 2x3 = −2    R1 → R2 2 6 −4 2 |4 1 0 −1 1 R =  1 0 −1 1 |5  R2 → R1 ≈  2 6 −4 2 −3 2 −2 0 |−2 −3 2 −2 0 →   → 1 |5 |4  R2 → −2R1 + R2 ≈  0 R3 → 3R1 + R3 0 |−2



1 0 −1 1 R =  2 6 −4 2 −3 2 −2 0 

1 R= 0 0

0 6 2

−1 −2 −5



0 1 2

−1 − 13 −5



0 1 0

−1 − 31 − 13 3



0 1 0

−1 − 13 1

1 R= 0 0 1 R= 0 0 1 R= 0 0

1 0 3

1 0 3

  1 → |5 |−6  R2 → R2 /6 ≈  0 → |13 0

0 1 2

  |5 1 → ≈ 0 → |−1  0 R3 → −2R2 + R3 |13

  → 1 1 |5 ≈ 0 → 0 |−1  3 R3 → − 13 0 R3 3 |15 1 0 9 − 13

0 1 0

4 20 4 20 − t x4 → − 13 13 13 13

x2 = −

3 28 3 28 x4 → − + t + 13 13 13 13

x3 = −

45 9 45 9 x4 → − + t + 13 13 13 13 x4 = t

1 0 3

 |5 |−6  |13

 −1 1 |5 −31 0 |−1  −5 3 |13 −1 − 13 − 133

0 1 0

−1 − 13 1

  1 |5 R1 → R! + R3  R2 → R2 + 1 R3 ≈  0 |−1  3 − 45 0 → 13 x1 =

−1 −2 −5

0 6 2

 |5 |4  |−2

0 1 0

1 0 9 − 13 0 0 1

1 0 3

 |5 |−1  |15  |5  |−1 − 45 13

4 13 3 − 13 9 − 13

 20    1328  −  13 − 45 13...


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