Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Dido525

  • 2 years ago

Integrals and linear algebra?

  • This Question is Closed
  1. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 Attachment
  2. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I can't figure out the damn matrix lol.

  3. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Please help.

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

    Are we missing part (a)?

  5. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay, I guess I should include that.

  6. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think it is just \([\frac{8}{3},2,2]\)

  7. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think so too. It's 5 marks though. Is it really that easy?

  8. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    because if you multiply that by \([a,b,c]\) written of course as a column, you get just what you want

  9. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Here is part a. I don't think it's required though.

    1 Attachment
  10. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmm, Feels too easy lol.

  11. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks!

  12. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it does seem easy, but i am fairly sure it works right?

  13. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah it does.

  14. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok

  15. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Can you answer another question similar to that?

  16. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i dunno my linear algebra is sometime weak, but i can try

  17. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 Attachment
  18. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @tkhunny

  19. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So f(0)=c f(1)=a+b+c f(2)=4a+2b+c

  20. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so: |dw:1359170790953:dw|

  21. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    right?

  22. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    How did you deduce that matrix?

  23. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i am a bit confused because those are \(f\) values not \(p\) values. \(f(0)=0 f(1)=? , f(2)=?\) are you supposed to use \(p\) instead of \(f\) ?

  24. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But we are given: |dw:1359171043668:dw|

  25. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So yeah, How did you deduce that? @tkhunny

  26. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    then i guess it is the same as the other one, because the integral is \(\frac{8}{3}a+2b+2\)

  27. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oops i meant \(\frac{8}{3}a+2b+2c\)

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

    Never mind that silly thing, I still haven't figured out what we're doing, here. I believe the premise is that we CANNOT find an antiderivative of f(x), but we CAN find one for p(x).

  29. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah, I guess because f(x) isn't given.

  30. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Any ideas?

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

    \(\int\limits_{0}^{2}p(x)\;dx = \int\limits_{0}^{2}ax^{2} + bx + c\;dx = \dfrac{8}{3}a + 2b + 2c \approx \int\limits_{0}^{2}f(x)\;dx\) I'm not sure what else we are looking at, here.

  32. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well I posted the question. It dosen't make sense to me though :( .

  33. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

    I see it. Sorry, it took me a minute. It will take a moment to code and it requires Part A!

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

    Part A \[A = \left[\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 1 \\ 4 & 2 & 1\end{matrix}\right]\] \[\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = A * \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\] \[A^{-1} = \dfrac{1}{4}\left[\begin{matrix}1 & -2 & 1 \\ -4 & 4 & 0 \\ 4 & 0 & 0\end{matrix}\right]\] \[A^{-1\cdot}\left(\begin{matrix}f(0) \\ f(1) \\ f(2)\end{matrix}\right) = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]

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

    Part B \[i = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\cdot \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]

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

    @Dido525 Are you SURE you don't want to do the glorious Part C? It's now staring at you!

  38. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry I was eating lol.

  39. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well I will attempt it.

  40. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well We know: |dw:1359173248519:dw|

  41. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    right?

  42. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @tkhunny

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

    No, \(A\) is no longer useful. Look at the very bottom of Part A and compare that carefully to Part B, right below it. Make the substitution and you are done. \(C = B\cdot A^{-1}\) I guess I forgot to say : \[B = \left(\begin{matrix}\dfrac{8}{3} & 2 & 2\end{matrix}\right)\]

  44. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So it would be: |dw:1359174107470:dw|

  45. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Like the inverse of A.

  46. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Which I have to find.

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

    No, that's \(B\cdot A\) you need \(B\cdot A^{-1}\).

  48. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah. I mean Once I find the inverse I set it up like that right?

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

    Yes, but for future reference, don't write \(A\) as a place holder for \(A^{-1}\). Very confusing.

  50. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Alright. Lets grind this out...

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

    You may find the result surprising. It's a bit, shall we say, regular!

  52. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2/3 2/3 2/3 ?

  53. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    for c?

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

    Big C, not little c, but yes. Are you surprised?

  55. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am actually! :P .

  56. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Can I ask something?

  57. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    How did you make that substitution?

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

    What a strange result. Sample the function at both ends and in the middle and then use 2/3 of those values and you've created a quadratic approximation. Too simple!

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

    You have to come to terms with the very last equation in my Part A. This gives you an expression for (a b c )^T in terms of f(0), f(1), and f(2). We are simply substituting this expression for (a b c)^T from Part B.

  60. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    And shoudn't the second row of matrix a say 1 1 1 ? Instead of 0 1 1.

  61. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Matrix A*

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

    Whoops! You are right. This changes things and that silly 2/3 things IS as weird as it looked. I like \(\left(\begin{matrix}\dfrac{1}{3} & \dfrac{4}{3} & \dfrac{1}{3}\end{matrix}\right)\) much better.

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

    You should get \(A^{-1} = \dfrac{1}{2}\left[\begin{matrix}1 & -2 & 1 \\ -3 & 4 & -1 \\ 2 & 0 & 0\end{matrix}\right]\) Important lesson, never mess up Matrix \(A\) when you will need it later!

  64. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep Yep, I did XD .

  65. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Except I didn't bother factoring out 1/2 .

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

    Perfect. The factoring just makes it simpler to write.

  67. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Still lost on the substitution... Or else I got you just fine.

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

    Just calculate \(B\cdot A^{-1}\)

  69. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I did. I got it correct and everything. I just don't get why that works.

  70. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    OOO!!!! I see it! Nvm.

  71. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks so much!!!

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

    Hmmm...\(\left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\) is the key. Stare at it until it soaks in. That's all I did while you were having dinner. :-)

  73. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep, you just substituted for that right?

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

    That's it, but I used Part A to develop the substitution. Do we remember someone saying Part A was not helpful? Good work! Way to hang in there.

  75. Dido525
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thank you so much! You helped me out immensely :) .

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

    • Attachments:

Ask your own question

Sign Up
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.