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KonradZuse

6.4 #2b In Exercise 2 find the least squares solution of the linear equation Ax = b. b. > A = (Matrix(3, 2, {(1, 1) = 2, (1, 2) = -2, (2, 1) = 1, (2, 2) = 1, (3, 1) = 3, (3, 2) = 1})); b = (Vector(3, {(1) = 2, (2) = -1, (3) = 1}));

  • one year ago
  • one year ago

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  1. KonradZuse
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    \[\left[\begin{matrix}2 & -2 \\ 1 & 1 \\ 3 & 1\end{matrix}\right]\] \[\left(\begin{matrix}2 \\ -1 \\ 1\end{matrix}\right)\]

    • one year ago
  2. KonradZuse
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    now my book says to use the formula A^T A = a^T b

    • one year ago
  3. KonradZuse
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    and then solve.... I'll post the pratice problem.

    • one year ago
  4. KonradZuse
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    • one year ago
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  5. tkhunny
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    You need a most marvelous construct: \(\left(A^{t}\cdot A\right)^{-1}\cdot A^{t}\) It is a beautifu thing.

    • one year ago
  6. KonradZuse
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    how do we get that solution of x1 and x2? Maybe I'm over tihnking it...

    • one year ago
  7. KonradZuse
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    Hmm... That's diff from the formula they say...

    • one year ago
  8. tkhunny
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    No, it is the same. It's just the "solved" version given the conclusion that \(\left(A^{t}\cdot A\right)\) is NonSingular.

    • one year ago
  9. KonradZuse
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    IUc... so how do I figure out x1 and x2 at the end? I'm just confused on that part... I understand how to do it if it was just 1 matrix, but a matrix * x = another matrix I'm not so sure about?

    • one year ago
  10. KonradZuse
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    like how did they get the answers to the question I posted above?

    • one year ago
  11. tkhunny
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    Construct that delightful matrix. I get [3/7,-2/3]^T Find \(A^{t}\cdot A\), first.

    • one year ago
  12. KonradZuse
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    The one I posted, or their example?

    • one year ago
  13. tkhunny
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    I didn't look at the piosted example. Let's go with your original problem statement.

    • one year ago
  14. KonradZuse
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    kk

    • one year ago
  15. KonradZuse
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    so we have \[\left[\begin{matrix}2 & -2 \\ 1 & 1 \\ 3 & 1\end{matrix}\right]\]

    • one year ago
  16. tkhunny
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    That's "A".

    • one year ago
  17. KonradZuse
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    yup now to find A^t

    • one year ago
  18. tkhunny
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    Which we hope is trivial, except for the type-setting. :-)

    • one year ago
  19. KonradZuse
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    A^T should be \[\left[\begin{matrix}2 & 1 & 3 \\ -2 & 1 & 1 \end{matrix}\right]\]

    • one year ago
  20. tkhunny
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    Multiply away and you will be pleasantly surprised at the result.

    • one year ago
  21. KonradZuse
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    \[\left[\begin{matrix}14 & 0 \\ 0 & 6\end{matrix}\right]\] is what I got.

    • one year ago
  22. tkhunny
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    Side note: Be a little more careful with your notation. So far, you have used "A" and "a' interchangeably. Also, both "T" and "t" for transpose. More consistent behavior will eliminate errors.

    • one year ago
  23. tkhunny
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    Perfect. Now the inverse of that.

    • one year ago
  24. KonradZuse
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    The book doesn't do that but okay.

    • one year ago
  25. tkhunny
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    Positive and Diagonal - Definitely NonSingular!

    • one year ago
  26. KonradZuse
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    \[\left[\begin{matrix}6 & 0 \\ 0 & 14 \end{matrix}\right]\]

    • one year ago
  27. tkhunny
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    Fair enough. I took a look at the example. They are setting up a small system of equestions and asking you to solve the system using any method at your disposal. On method of solving such a system is hte Matrix Inverse. That is what we are doing.

    • one year ago
  28. tkhunny
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    Close! Divide by 84.

    • one year ago
  29. KonradZuse
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    I forgot the 1/(ad- bc)

    • one year ago
  30. tkhunny
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    Remember that \(A\cdot A^{-1} = I\)

    • one year ago
  31. tkhunny
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    Yes, that's all you left off. Good work.

    • one year ago
  32. KonradZuse
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    1/(16 * 4) - (0 * 0)

    • one year ago
  33. KonradZuse
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    = 1/64

    • one year ago
  34. tkhunny
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    84

    • one year ago
  35. KonradZuse
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    woop sI did 4 not 6 :p

    • one year ago
  36. KonradZuse
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    and 16 not 14 lol...

    • one year ago
  37. KonradZuse
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    yeah 1/84

    • one year ago
  38. KonradZuse
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    so we don't have to take the inverse, do we?

    • one year ago
  39. tkhunny
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    ahahah Yes, \(16*4 \ne 14*6\)

    • one year ago
  40. tkhunny
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    You did already find the inverse. It is 1/84 times the 6,14 diagonal matrix.

    • one year ago
  41. KonradZuse
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    yeah but since the book doesn't, do we have to?

    • one year ago
  42. tkhunny
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    Yes, the book does. When it introduces x1 and x2, that is exactly what it is doing. The book just isn't using matrix methods to solve the system. This strikes me as rather odd, since this is inherently a matrix problem.

    • one year ago
  43. KonradZuse
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    hmmm... Konfused... but if you say so ;P.

    • one year ago
  44. tkhunny
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    You started with Ax = b We added the transpose: \(A^{t}\cdot A\cdot x = A^{t}\cdot b\)

    • one year ago
  45. KonradZuse
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    mhm.

    • one year ago
  46. tkhunny
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    Now we have a system to solve. One way to solve this is to find hte inverse of \(A^{t}\cdot A\), if it has an inverse. It does.

    • one year ago
  47. KonradZuse
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    \[\left[\begin{matrix}\frac{1}{14} & 0 \\ 0 & \frac{1}{6}\end{matrix}\right]\]

    • one year ago
  48. tkhunny
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    This gives: \((A^{t}\cdot A)^{-1}\cdot (A^{t}\cdot A)\cdot x = (A^{t}\cdot A)^{-1}\cdot A^{t}\cdot b\), using the inverse to solve the system.

    • one year ago
  49. KonradZuse
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    seems like we are just adding it in :P

    • one year ago
  50. KonradZuse
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    shall I solve for A^T B?

    • one year ago
  51. KonradZuse
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    and solving the first portion yeilds the identity matrix.

    • one year ago
  52. tkhunny
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    The inverse SOLVES the system, just like substitution or elimination or whatever other method you might use to solve the system.

    • one year ago
  53. tkhunny
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    Yes, that is what hte inverse does. The left-hand-side is just 'x' when we are done. This symbolizes the solved system.

    • one year ago
  54. KonradZuse
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    I think I'm understanding where this is going... Instead of having to row reduce....

    • one year ago
  55. KonradZuse
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    lke a billion times easier :P

    • one year ago
  56. KonradZuse
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    I think I've been doing it the hard way :p

    • one year ago
  57. KonradZuse
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    \[\left(\begin{matrix}6 \\ -4\end{matrix}\right)\]

    • one year ago
  58. KonradZuse
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    A^Tb

    • one year ago
  59. tkhunny
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    Different problems may benefit from different methods. I think you're not quite done with this one. We don't need just \(A^{t}\cdot b\), we need \(x = (A^{t}\cdot A)^{-1}\cdot A^{t}\cdot b\)

    • one year ago
  60. KonradZuse
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    that * the inverse = \[\left(\begin{matrix}\frac{3}{7} \\ \frac{-2}{3}\end{matrix}\right)\]

    • one year ago
  61. tkhunny
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    Done! VERY good work. This was cutting-edge stuff 40 years ago. That's WAY closer to current research than your basic 300-400 year old calculus. :-)

    • one year ago
  62. KonradZuse
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    so x1 = 3/7 and x2 = -2/3???

    • one year ago
  63. KonradZuse
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    or am I supposed to solve for that now?!

    • one year ago
  64. tkhunny
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    That's it.

    • one year ago
  65. tkhunny
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    Just for the record, just this Friday, I did a job for my employer using exactly this technique. My "A" was (462 x 4), so I was pretty glad I didn't have to do it by hand. :-)

    • one year ago
  66. KonradZuse
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    Damn.... Well that was easier than what the book was doing if it came out all nicely like that :) Thanks!

    • one year ago
  67. tkhunny
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    Like I said, it is a little odd that they would introduce this technique and just bail at the end, requiring you to resort to Algebra II. Maybe the author thinks you're not ready for the Inverse? Odd, indeed.

    • one year ago
  68. KonradZuse
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    This book really SUCKS..... Sometimes it's okay....

    • one year ago
  69. KonradZuse
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    I have another Q that no one really answered if you don't mind taking a look at it.. I just wanted to be sure I did it correctly.

    • one year ago
  70. tkhunny
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    Well, good luck with THAT (the book and author, that is)! (laughing)

    • one year ago
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