A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Need help with Applied Linear Algebra

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I found the reduced echelon form to be : |dw:1443287325608:dw|

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and now I do not know what to do, So I know that there will never be a "No Solution" to this system. and I know there will be a unique solution when a=0 (Correct me if I am wrong) but when will there be infinite solution?

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ganeshie8

  5. surry99
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Assuming your reduced matrix is correct, the only solution would be a=0

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so it will just be a trivial solution

  7. surry99
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but what if I just have as a row echelon form like this: |dw:1443288786467:dw| the answer would change then, then "a" will have to equal -1 for their to be a trivial solution

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    should i use row echelon form or reduced form?

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\left|\begin{matrix}1 & -1 &4 \\ 3 & -2 &10\\a&3&-8\end{matrix}\right|=1[\left( -2 \right)\left( -8 \right)-10*3]-(-1)[3*(-8)-10a]+4[3*3+2a]\] =16-30-24-10a+36+8a =-2-2a it has infinite solutions if-2-2a=0 or a=-1 no solution if \[a \neq-1\] only trivial solution x1=x2=x3=0

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but homogenous systems never have "no solutions". so when it asks when It has no solution I would have to write "never". wait that doesnt make sense, if a=-1 I dont think there will be a inifinite solutions because then (-2-2a)x3=0 so x3 must equal 0. making all the other variables(x1, x2, x3=0) so there will be only 1 solution not infinite right?

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so ultimately when it asks when does the system have unique solution I would have to write when a=-1. and when it asks when does it have infinite solution, I would have to write "never". is that correct?

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you are correct if ax=0 then either a=0 or x=0

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.