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 | |
Total Downloads | 27 |
Total Views | 138 |
Ejercicios de Sistemas Consistentes e Inconsistentes...
−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...