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KonradZuse

  • 2 years ago

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}));

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

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

  3. KonradZuse
    • 2 years ago
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    and then solve.... I'll post the pratice problem.

  4. KonradZuse
    • 2 years ago
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  5. tkhunny
    • 2 years ago
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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.

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

  7. KonradZuse
    • 2 years ago
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    Hmm... That's diff from the formula they say...

  8. tkhunny
    • 2 years ago
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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.

  9. KonradZuse
    • 2 years ago
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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?

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

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

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

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

  14. KonradZuse
    • 2 years ago
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    kk

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

  16. tkhunny
    • 2 years ago
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    That's "A".

  17. KonradZuse
    • 2 years ago
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    yup now to find A^t

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

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

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

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

  22. tkhunny
    • 2 years ago
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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.

  23. tkhunny
    • 2 years ago
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    Perfect. Now the inverse of that.

  24. KonradZuse
    • 2 years ago
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    The book doesn't do that but okay.

  25. tkhunny
    • 2 years ago
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    Positive and Diagonal - Definitely NonSingular!

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

  27. tkhunny
    • 2 years ago
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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.

  28. tkhunny
    • 2 years ago
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    Close! Divide by 84.

  29. KonradZuse
    • 2 years ago
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    I forgot the 1/(ad- bc)

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

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

  32. KonradZuse
    • 2 years ago
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    1/(16 * 4) - (0 * 0)

  33. KonradZuse
    • 2 years ago
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    = 1/64

  34. tkhunny
    • 2 years ago
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    84

  35. KonradZuse
    • 2 years ago
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    woop sI did 4 not 6 :p

  36. KonradZuse
    • 2 years ago
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    and 16 not 14 lol...

  37. KonradZuse
    • 2 years ago
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    yeah 1/84

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

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

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

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

  42. tkhunny
    • 2 years ago
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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.

  43. KonradZuse
    • 2 years ago
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    hmmm... Konfused... but if you say so ;P.

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

  45. KonradZuse
    • 2 years ago
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    mhm.

  46. tkhunny
    • 2 years ago
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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.

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

  48. tkhunny
    • 2 years ago
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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.

  49. KonradZuse
    • 2 years ago
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    seems like we are just adding it in :P

  50. KonradZuse
    • 2 years ago
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    shall I solve for A^T B?

  51. KonradZuse
    • 2 years ago
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    and solving the first portion yeilds the identity matrix.

  52. tkhunny
    • 2 years ago
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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.

  53. tkhunny
    • 2 years ago
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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.

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

  55. KonradZuse
    • 2 years ago
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    lke a billion times easier :P

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

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

  58. KonradZuse
    • 2 years ago
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    A^Tb

  59. tkhunny
    • 2 years ago
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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 \(x = (A^{t}\cdot A)^{-1}\cdot A^{t}\cdot b\)

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

  61. tkhunny
    • 2 years ago
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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. :-)

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

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

  64. tkhunny
    • 2 years ago
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    That's it.

  65. tkhunny
    • 2 years ago
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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. :-)

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

  67. tkhunny
    • 2 years ago
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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.

  68. KonradZuse
    • 2 years ago
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    This book really SUCKS..... Sometimes it's okay....

  69. KonradZuse
    • 2 years ago
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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.

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

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