## angela210793 Group Title Find all the soulutions possible ... 2 years ago 2 years ago

1. angela210793 Group Title

using matrices |dw:1333988132875:dw|

2. angela210793 Group Title

since |A"=0 can qe use KRONECKER-CAPELLI THEOREM i did smth like |dw:1333988210313:dw|

3. angela210793 Group Title

|dw:1333988330791:dw| but latter using gauss i got |dw:1333988378631:dw|

4. angela210793 Group Title

did i do smth wrong?? |dw:1333988426922:dw|

5. experimentX Group Title

seems like the first and the last plane are parallel.

6. angela210793 Group Title

7. experimentX Group Title

when ever two planes are parallel, they do not have solution.

8. experimentX Group Title

first and the last.

9. experimentX Group Title

and two planes will give you infinite solution.

10. experimentX Group Title

a line to be exact.

11. angela210793 Group Title

wht will b the soluyions??? can u find them Experiment?

12. angela210793 Group Title

solutions****

13. eliassaab Group Title

k=1, the last equation will be -2 times the first one The solutions lie on the line of intersections of the last two plane.

14. experimentX Group Title

you will have infinite solutions, substitute z from second and third.

15. experimentX Group Title

that will be a line, and line has infinite points. even if the first and the last line were not parallel, you were going to have infinite solution, because 4 variables and 3 equations---> that usually ends up that way.

16. angela210793 Group Title

noo..k is a costant

17. experimentX Group Title

because you are free to choose any value of k, ... and for particular value of k, you will have particular solution.

18. experimentX Group Title

your solution WOULD have been in terms of k (that what i meant by infinite).

19. angela210793 Group Title

lemme think a lil bit please

20. angela210793 Group Title

tht's wht i did by using Kroneker-capelli....but the prob is that i get 0+0+0=2 at the last equation.... maybe i am being too dumb but i am really not getting it

21. experimentX Group Title

says that A must have same rank as [A|b]

22. experimentX Group Title

23. angela210793 Group Title

yes...since r(A)=2 r(Ab) must be 2 too right

24. experimentX Group Title

A has rank 2 i guess

25. angela210793 Group Title

what language is that Experiment??? O.o

26. experimentX Group Title

it has rank two ... checked from soft. Croatian ... i don't understand too. i used google to translate.

27. experimentX Group Title

can you tell me what is [A|b] ??

28. angela210793 Group Title

r(Ab)=4-4k at least tht's wht i found

29. angela210793 Group Title

|dw:1333989679050:dw|

30. angela210793 Group Title

r u getting anything?

31. experimentX Group Title

Nop .. I am not really quite into mathematics ... I am just experimental

32. angela210793 Group Title

hmmm.....ok...:) Thanks anyway :D I appreciate it :))

33. experimentX Group Title

i couldn't understand what does [A|b] represent.

34. angela210793 Group Title

Ab is the matrice formed by the constants before variables and the solutions |dw:1333990018496:dw|

35. angela210793 Group Title

not soultions sorry..they r not called solutions..I'm just not finding the right word -_-

36. experimentX Group Title

oh .. yeah i remember doingnthat (thanks to you). let's see what can i do for any arbitrary value .

37. angela210793 Group Title

@hoblos any idea???

38. experimentX Group Title

for k=5, we get rank 3

39. experimentX Group Title

okay i will check this value from -1000 to 1000

40. angela210793 Group Title

no actually u get rang 3 for every k different from 1

41. experimentX Group Title

it seems that we get rank 2 for k = -1

42. angela210793 Group Title

oh wait...did i go wrong by equaling 4-4k=2?????????? O.o...i think i shouldn't have done that

43. experimentX Group Title

I think i understand what's going on.

44. experimentX Group Title

LOL ... this is quite funny.

45. FoolForMath Group Title

I have never heard of KRONECKER-CAPELLI THEOREM.

46. experimentX Group Title

it's amazing to know that @ffm does not know something. http://lavica.fesb.hr/mat1/predavanja/node38.html

47. angela210793 Group Title

my google doesn't translate tht -_-

48. eliassaab Group Title

$x = \frac 3 2 -z \quad y = \frac 1 2$ z can be any number

49. experimentX Group Title

use translate feature of chrome.

50. angela210793 Group Title

Can u show me how u got that please @eliassaab ???

51. eliassaab Group Title

write x + y = 2 - z -2 x + 2 y = 2 z - 2 and solve the two equations with repect to x and y

52. hoblos Group Title

|dw:1333990759995:dw| this can be reduced to |dw:1333990839175:dw| to find solutions.. r(A) must be same as r(A|B) so -2+2k=0 ----> k=1 thus 2y=1 ---> y=1/2 x-1/2 + z = 1 x+z = 3/2

53. eliassaab Group Title

x + y = 2 - z -x + y = z - 1 add the two equation to get y=1/2 and then find x

54. hoblos Group Title

sorry... instead of (-2-2k) it is (-2+2k)

55. angela210793 Group Title

But Hoblos don't u get |dw:1333991218301:dw|

56. angela210793 Group Title

i got -2-2k O.o cause i tried this too

57. hoblos Group Title

mm i cant see the whole drawing!! always different from what?

58. angela210793 Group Title

zero

59. angela210793 Group Title

u know what....take the system and try solving it urself and ignore everything i've said

60. hoblos Group Title

it must be equals to zero in order to have solutions!

61. experimentX Group Title

for other values of k other than -1, we will have two parallel planes and one non parallel. means two lines. for k=-1, we have first plane = last pane, so we will have one line (I think this is what the theorem is trying to tell .... I did not understand anything)

62. angela210793 Group Title

i think the solutions r depended on k for k=1 are the soluions tht u find wht abt for kdifferent from 1?/

63. FoolForMath Group Title

Can't we use the rank method?

64. hoblos Group Title

for k different than 1 we dont have solutions..

65. experimentX Group Title

@hoblos it's -1 i think

66. angela210793 Group Title

tht's wht we're doing Fool...Kronecker-capelli :P u don't have solutions??? how??????

67. eliassaab Group Title

If $k\ne 1$, there is no solutions. If $k= 1$, there are infinite number of solutions that I listed before,

68. hoblos Group Title

why would it be -1 @experimentX ??!!

69. FoolForMath Group Title

Oh god! I only didn't knew the name

70. angela210793 Group Title

@FoolForMath tht makes Anxhela happy :) now start solving it....I'm going crazyyyy!!!!!!!!! :P

71. experimentX Group Title

because when k=1, we will have two lines ... for system of three planes.

72. FoolForMath Group Title

But then I am not a Russian :P So I guess I can't blame my teacher

73. FoolForMath Group Title
74. angela210793 Group Title

lol :D

75. hoblos Group Title

well.. considering the first equation and the last one.. x-y+z =k *(-2) -2x+2y-2z=-2 -2x+2y-2z=-2k -2x+2y-2z=-2 so k must be 1

76. eliassaab Group Title

You really do not need heavy Theorems to do a rather simple problem.

77. experimentX Group Title

i guess .. so i might have made computation mistake.

78. angela210793 Group Title

@eliassaab this is how we solve it in my univ..... @experimentX @hoblos @FoolForMath @eliassaab thanks a lot for trying and willing to help me :) forget this question :) I'll think again and repost it :))) thanks :)

79. experimentX Group Title

damn;; my mistake was computational ... LOL A = [1, -1, 1, i; 1, 1, 1, 2; -2, 2, -2, 2]; should have been -2 at the end.

80. hoblos Group Title

i remembred something we used to do... when r(A|B) < number of variables... we have no exact solutions.. instead we use a parameter so we take z=m this gives x= 3/2 -m so the solution is (3/2 -m, 1/2 , m) for any m belong to R