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Γραmmική ΄Αλγεβρα Θεmελιώδεις Θεώρηmα (mέρος 1ο) Πανεπιστήmιο Θεσσαλίας 15 Dεκεmβρίου 2013

# 22η Διάλεξη - Θεμελειώδες Θεώρημα Γραμμικής Άλγεβρας (μέρος 1ο)

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### Text of 22η Διάλεξη - Θεμελειώδες Θεώρημα Γραμμικής Άλγεβρας...

• 1. ( 1) 15 2013

2. A L1 U. 3. A L1 U. . A = LU L1 A = U 4. A L1 U. . A = LU L1 A = U 5. - 1 - A U . 6. - 12 - A U . - A U . 7. - 123 - A U . - A U . - ( ). 8. - 1234 - A U . - A U . - ( ). - L1 U. 9. n. . 10. n. . m. 11. n. . m. . . 12. 13. B A BA = I 14. B A BA = I C A AC = I 15. B A BA = I C A AC = I n = m = r . 16. B = AT A1AT 17. B = AT A1C = AT AATAT1 18. B = AT A1C = AT AATAT1 ( ) AT A AAT ; 19. 20. Ax = b b A Rm 21. Ax = b b A Rm Ax = b b r = m ( m n) 22. Ax = b b A Rm Ax = b b r = m ( m n) r = m ( m n) A 23. Ax = b b A Rm Ax = b b r = m ( m n) r = m ( m n) A 24. 25. Ax = b b A 26. Ax = b b A Ax = b b r = n ( n m) 27. Ax = b b A Ax = b b r = n ( n m) r = n ( n m) A 28. Ax = b b A Ax = b b r = n ( n m) r = n ( n m) A 29. 30. Ax = b x = Cb 31. Ax = b x = Cb Ax = ACb = b, 32. Ax = b x = Cb Ax = ACb = b, (C ). 33. Ax = b x = Cb Ax = ACb = b, (C ). Ax = b x = BAx = Bb . 34. A=4 0 0 0 5 0 35. A=4 0 0 0 5 0m = 2, n = 3, r = 2 36. 1 2 1 1 4 2 2 A= 8 4 4 2 1 1 37. 1 2 1 1 1 4 2 2 2 = 2 1 1 A= 8 4 4 4 2 1 1 1 38. 1 2 1 1 1 4 2 2 2 = 2 1 1 A= 8 4 4 4 2 1 1 1 A 1 A = uv T

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