A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

et A and B be 2*2 matrices such that 2A+3B= 1 2 -1 -2 Atranspose +B transpose=-1 3 -7 6 Compute A and B

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

    one sec I'll do it real quick

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

    maybe not quickly

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

    :D take ur time:)

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

    ooooo right bad at addition, woops

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

    u can attach a picture of ur paper with solution

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

    check to see if these work correctly: A = |-4 -23| |16 -5|

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

    B = |3 26| |-7 8|

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

    may u send me ur steps also

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

    nono B = |3 16| |-7 8|

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

    umm it's fairly illegible

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

    just attach a picture with ur steps and btw its perfectly right:)

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

    the camera on my phone is pretty terrible, can't read my writing in pencil

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

    steps I used: A^T + B^T = (A+B)^T thus: a11 + b11 = -1 a22+ b22 = 6 a12 + b12 = -7 a21 + b21 = 3 and i solved for the a values and plugged them into the other set of equations

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

    which are: 2a11 + 2b11 = 1 2a12 + 3b12 = 2 2a21 + 3b21 = -1 2a22+ 3b22 = -2

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

    which gave me the values for B matrix then I plugged those values back into the first set of equations to get my A matrix

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

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

    thanks alot:)

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

    yep, matricies are fun!

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

    wants alot of them?:D

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

    hit me with another one, I'm tired of my linear algebra studying anyway

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

    Let A be a 2*2 matrix Prove that if det(a)not equal zero then the inverse is given by A^-1=(1/det(A))adj(A)

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

    check that alsoo s

    1 Attachment
  23. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm, i konw it has to do with the cofactor expansion, let me chew on that a sec

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

    err, cofactor matrix

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

    ??

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

    thinking on it, I should really know this.

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

    ok :) take ur time

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

    from cramer's rule we get:\[x_{j} = \det(A _{i}(e _{j})) / \det(A)\] thus \[x _{ij} = (1/\det(A))C _{ji}\] which gives us:\[A^{-1} = X = (1/\det(A))[C _{ji}] = (1/\det(A))[C _{ij}]^T\]

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

    and \[[C _{ij}]^T = adj(A)\]

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

    might need an Assume the \[\det(A) \ne 0\]

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

    ??

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

    i pieced together a few statements, I'm pretty sure that is the right aproach but can't tell you how to do it exactly

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

    and about the attachment did u check it? may u please check it:)

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

    couldn't load it

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

    http://en.wikipedia.org/wiki/Cramer's_rule

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

    thanks:) i will send u the attachment again

    1 Attachment
  37. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

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

    you there?

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

    yeah sorry lost the connection

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

    thats the Gauss Jordan algorithm metod whats about the adjugates

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

    are u there?

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

    wouldn't let me type, look at both

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

    perfect:)

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

    want more?

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

    sure

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

    start a new thread though

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

    1 Attachment
  49. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohh sorry ok:)

  50. 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.