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angela210793

  • 2 years ago

Find all the soulutions possible ...

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  1. angela210793
    • 2 years ago
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    using matrices |dw:1333988132875:dw|

  2. angela210793
    • 2 years ago
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    since |A"=0 can qe use KRONECKER-CAPELLI THEOREM i did smth like |dw:1333988210313:dw|

  3. angela210793
    • 2 years ago
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    |dw:1333988330791:dw| but latter using gauss i got |dw:1333988378631:dw|

  4. angela210793
    • 2 years ago
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    did i do smth wrong?? |dw:1333988426922:dw|

  5. experimentX
    • 2 years ago
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    seems like the first and the last plane are parallel.

  6. angela210793
    • 2 years ago
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    which???? show me please

  7. experimentX
    • 2 years ago
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    when ever two planes are parallel, they do not have solution.

  8. experimentX
    • 2 years ago
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    first and the last.

  9. experimentX
    • 2 years ago
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    and two planes will give you infinite solution.

  10. experimentX
    • 2 years ago
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    a line to be exact.

  11. angela210793
    • 2 years ago
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    wht will b the soluyions??? can u find them Experiment?

  12. angela210793
    • 2 years ago
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    solutions****

  13. eliassaab
    • 2 years ago
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    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
    • 2 years ago
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    you will have infinite solutions, substitute z from second and third.

  15. experimentX
    • 2 years ago
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    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
    • 2 years ago
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    noo..k is a costant

  17. experimentX
    • 2 years ago
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    because you are free to choose any value of k, ... and for particular value of k, you will have particular solution.

  18. experimentX
    • 2 years ago
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    your solution WOULD have been in terms of k (that what i meant by infinite).

  19. angela210793
    • 2 years ago
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    lemme think a lil bit please

  20. angela210793
    • 2 years ago
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    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
    • 2 years ago
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    says that A must have same rank as [A|b]

  22. experimentX
    • 2 years ago
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    use google translate http://lavica.fesb.hr/mat1/predavanja/node38.html

  23. angela210793
    • 2 years ago
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    yes...since r(A)=2 r(Ab) must be 2 too right

  24. experimentX
    • 2 years ago
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    A has rank 2 i guess

  25. angela210793
    • 2 years ago
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    what language is that Experiment??? O.o

  26. experimentX
    • 2 years ago
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    it has rank two ... checked from soft. Croatian ... i don't understand too. i used google to translate.

  27. experimentX
    • 2 years ago
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    can you tell me what is [A|b] ??

  28. angela210793
    • 2 years ago
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    r(Ab)=4-4k at least tht's wht i found

  29. angela210793
    • 2 years ago
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    |dw:1333989679050:dw|

  30. angela210793
    • 2 years ago
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    r u getting anything?

  31. experimentX
    • 2 years ago
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    Nop .. I am not really quite into mathematics ... I am just experimental

  32. angela210793
    • 2 years ago
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    hmmm.....ok...:) Thanks anyway :D I appreciate it :))

  33. experimentX
    • 2 years ago
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    i couldn't understand what does [A|b] represent.

  34. angela210793
    • 2 years ago
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    Ab is the matrice formed by the constants before variables and the solutions |dw:1333990018496:dw|

  35. angela210793
    • 2 years ago
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    not soultions sorry..they r not called solutions..I'm just not finding the right word -_-

  36. experimentX
    • 2 years ago
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    oh .. yeah i remember doingnthat (thanks to you). let's see what can i do for any arbitrary value .

  37. angela210793
    • 2 years ago
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    @hoblos any idea???

  38. experimentX
    • 2 years ago
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    for k=5, we get rank 3

  39. experimentX
    • 2 years ago
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    okay i will check this value from -1000 to 1000

  40. angela210793
    • 2 years ago
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    no actually u get rang 3 for every k different from 1

  41. experimentX
    • 2 years ago
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    it seems that we get rank 2 for k = -1

  42. angela210793
    • 2 years ago
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    oh wait...did i go wrong by equaling 4-4k=2?????????? O.o...i think i shouldn't have done that

  43. experimentX
    • 2 years ago
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    I think i understand what's going on.

  44. experimentX
    • 2 years ago
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    LOL ... this is quite funny.

  45. FoolForMath
    • 2 years ago
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    I have never heard of KRONECKER-CAPELLI THEOREM.

  46. experimentX
    • 2 years ago
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    it's amazing to know that @ffm does not know something. http://lavica.fesb.hr/mat1/predavanja/node38.html

  47. angela210793
    • 2 years ago
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    my google doesn't translate tht -_-

  48. eliassaab
    • 2 years ago
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    \[ x = \frac 3 2 -z \quad y = \frac 1 2 \] z can be any number

  49. experimentX
    • 2 years ago
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    use translate feature of chrome.

  50. angela210793
    • 2 years ago
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    Can u show me how u got that please @eliassaab ???

  51. eliassaab
    • 2 years ago
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    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
    • 2 years ago
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    |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
    • 2 years ago
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    x + y = 2 - z -x + y = z - 1 add the two equation to get y=1/2 and then find x

  54. hoblos
    • 2 years ago
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    sorry... instead of (-2-2k) it is (-2+2k)

  55. angela210793
    • 2 years ago
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    But Hoblos don't u get |dw:1333991218301:dw|

  56. angela210793
    • 2 years ago
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    i got -2-2k O.o cause i tried this too

  57. hoblos
    • 2 years ago
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    mm i cant see the whole drawing!! always different from what?

  58. angela210793
    • 2 years ago
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    zero

  59. angela210793
    • 2 years ago
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    u know what....take the system and try solving it urself and ignore everything i've said

  60. hoblos
    • 2 years ago
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    it must be equals to zero in order to have solutions!

  61. experimentX
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Can't we use the rank method?

  64. hoblos
    • 2 years ago
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    for k different than 1 we dont have solutions..

  65. experimentX
    • 2 years ago
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    @hoblos it's -1 i think

  66. angela210793
    • 2 years ago
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    tht's wht we're doing Fool...Kronecker-capelli :P u don't have solutions??? how??????

  67. eliassaab
    • 2 years ago
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    If \[ k\ne 1 \], there is no solutions. If \[ k= 1 \], there are infinite number of solutions that I listed before,

  68. hoblos
    • 2 years ago
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    why would it be -1 @experimentX ??!!

  69. FoolForMath
    • 2 years ago
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    Oh god! I only didn't knew the name

  70. angela210793
    • 2 years ago
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    @FoolForMath tht makes Anxhela happy :) now start solving it....I'm going crazyyyy!!!!!!!!! :P

  71. experimentX
    • 2 years ago
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    because when k=1, we will have two lines ... for system of three planes.

  72. FoolForMath
    • 2 years ago
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    But then I am not a Russian :P So I guess I can't blame my teacher

  73. FoolForMath
    • 2 years ago
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    http://en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem

  74. angela210793
    • 2 years ago
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    lol :D

  75. hoblos
    • 2 years ago
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    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
    • 2 years ago
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    You really do not need heavy Theorems to do a rather simple problem.

  77. experimentX
    • 2 years ago
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    i guess .. so i might have made computation mistake.

  78. angela210793
    • 2 years ago
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    @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
    • 2 years ago
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    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
    • 2 years ago
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    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

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