Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

angela210793

  • 2 years ago

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

  • This Question is Closed
  1. wasiqss
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    system is hidden LOl

  2. Kreshnik
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait a second.. she must be writing it !!

  3. angela210793
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1334243246880:dw|

  4. wasiqss
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    anxhela , you want it to be done by cramer's rule or gauss elimination

  5. angela210793
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i did |dw:1334243367057:dw| i know wassiq...but u can't use cramer when determinative equals 0

  6. angela210793
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1334243466671:dw|

  7. wasiqss
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ahhh yes, determinant is zero . we must go for gauss elimination

  8. wasiqss
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do you know gauss, i can do that

  9. angela210793
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i did |dw:1334243615536:dw| krocker-capelli but i got

  10. angela210793
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i know Gauss but we have a k in there

  11. wasiqss
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good :), i havnt done krocker one til yet, but yes k would have been a headache in gauss

  12. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    k has to be 1

  13. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    if you want a solution

  14. angela210793
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how did u find it

  15. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Zarkon is the only one I won't chastise for just giving answer since they are so provocative ;)

  16. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    -2 times row one = row 3

  17. wasiqss
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Zarkon if k=1 , then game is over :D

  18. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    no...if \(k\ne 1\) then there is no soluition.

  19. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    solution

  20. angela210793
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    u see..i tried Gauss..and i got at the row3 0 0 0 2 where did i got wrong O.o

  21. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    \[\left.\begin{matrix}x-y+z=k \\ -2x+2y-2z=-2\end{matrix}\right.\] 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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    NIce work zarkon

  23. phi
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok...thanks

  27. phi
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  28. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.