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angela210793

  • 4 years ago

Find all the solutions possible of the system ...using matrices

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  1. wasiqss
    • 4 years ago
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    system is hidden LOl

  2. Kreshnik
    • 4 years ago
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    wait a second.. she must be writing it !!

  3. angela210793
    • 4 years ago
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    |dw:1334243246880:dw|

  4. wasiqss
    • 4 years ago
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    anxhela , you want it to be done by cramer's rule or gauss elimination

  5. angela210793
    • 4 years ago
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    i did |dw:1334243367057:dw| i know wassiq...but u can't use cramer when determinative equals 0

  6. angela210793
    • 4 years ago
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    |dw:1334243466671:dw|

  7. wasiqss
    • 4 years ago
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    ahhh yes, determinant is zero . we must go for gauss elimination

  8. wasiqss
    • 4 years ago
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    do you know gauss, i can do that

  9. angela210793
    • 4 years ago
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    i did |dw:1334243615536:dw| krocker-capelli but i got

  10. angela210793
    • 4 years ago
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    i know Gauss but we have a k in there

  11. wasiqss
    • 4 years ago
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    good :), i havnt done krocker one til yet, but yes k would have been a headache in gauss

  12. Zarkon
    • 4 years ago
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    k has to be 1

  13. Zarkon
    • 4 years ago
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    if you want a solution

  14. angela210793
    • 4 years ago
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    how did u find it

  15. TuringTest
    • 4 years ago
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    Zarkon is the only one I won't chastise for just giving answer since they are so provocative ;)

  16. Zarkon
    • 4 years ago
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    -2 times row one = row 3

  17. wasiqss
    • 4 years ago
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    Zarkon if k=1 , then game is over :D

  18. Zarkon
    • 4 years ago
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    no...if \(k\ne 1\) then there is no soluition.

  19. Zarkon
    • 4 years ago
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    solution

  20. angela210793
    • 4 years ago
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    u see..i tried Gauss..and i got at the row3 0 0 0 2 where did i got wrong O.o

  21. Zarkon
    • 4 years ago
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    \[\left.\begin{matrix}x-y+z=k \\ -2x+2y-2z=-2\end{matrix}\right.\] multiply first by 2 \[\Rightarrow\] \[\left.\begin{matrix}2x-2y+2z=2k \\ -2x+2y-2z=-2\end{matrix}\right.\] add you get \(0=2k-2\Rightarrow k=1\)

  22. wasiqss
    • 4 years ago
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    NIce work zarkon

  23. phi
    • 4 years ago
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    If you put the augmented matrix in reduced row echelon form you get \[\left[\begin{matrix}1 &0 & 1&1 +k/2 \\ 0 & 1 &0 & 1-k/2 \\ 0 & 0 & 0&2k-2 \\\end{matrix}\right]\]

  24. phi
    • 4 years ago
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    as zarkon noted, k must be 1 for there to be a solution. However, note that there are an infinite # of sols if k=1, because you have a non-zero null space

  25. angela210793
    • 4 years ago
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    the system would have infinite solutions for r(A)=r(Ab) and no solutions for r(A)<r(Ab) how do u find the range of Ab when u have k in there..r(A)=r(Ab) if k wasn't 1....sorry mozilla froze -_-

  26. angela210793
    • 4 years ago
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    ok...thanks

  27. phi
    • 4 years ago
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    The complete solution is \[\left(\begin{matrix}3/2 \\ 1/2 \\ 0\end{matrix}\right)+c\left(\begin{matrix}1 \\ 0 \\ -1\end{matrix}\right)\] where c is arbitrary

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