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KonradZuse

Given the linear system Ax = b with (a) Find all least squares solutions of the linear system Ax = b. (b) What is the normal system for this least squares problem? (c) Compute the least squares error to three decimal places.

  • one year ago
  • one year ago

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  1. KonradZuse
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    \[A =\left[\begin{matrix}1 & 3 \\ 3 & 9\\ -2 & -6\end{matrix}\right]\] \[B = \left(\begin{matrix}1 \\ 1\\ 0\end{matrix}\right)\]

    • one year ago
  2. KonradZuse
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    @UnkleRhaukus @Zarkon

    • one year ago
  3. KonradZuse
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    Since my work was deleted I will explain that I got up to the end where I need to solve, I am confused exactly how to solve the matrix [x1x2] col matrix = [4...12] which was my col matrix answer... Flipping out trying to get this done by midnight.... :(

    • one year ago
  4. KonradZuse
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    There was also a trick someone taught me that if I take the A * A^-1 I can use that to find my answer; however i got 0 for my determinant so........ Idk how that works, everything = 0? I don't think so...... Maybe?

    • one year ago
  5. KonradZuse
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    nope, it's correct.....

    • one year ago
  6. KonradZuse
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    Tk have you come to save me? :P

    • one year ago
  7. tkhunny
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    Ax = b Introduce the Transpose \(A^{T}Ax = A^{T}b\) Hope that \(A^{T}A\) is non-singular and calculate it's inverse: \(\left(A^{T}A\right)^{-1}\left(A^{T}A\right)x = \left(A^{T}A\right)^{-1}A^{T}b\) \(Ix = \left(A^{T}A\right)^{-1}A^{T}b\)

    • one year ago
  8. KonradZuse
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    Oh were you the one who showed me that trick with the -1? When taking the determinant I get 0... What happens then? Is it all 0? lol

    • one year ago
  9. tkhunny
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    Determinant? It's an inverse.

    • one year ago
  10. tkhunny
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    Unfortunately, your \(\left(A^{T}A\right)^{-1}\) is singular. We should have seen this from the original data. Column 2 is rather obviuosly twice Column 1. Where does that leave us?

    • one year ago
  11. KonradZuse
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    I thought taking the inverse is the same as taking the determinant?

    • one year ago
  12. KonradZuse
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    I guess I konfused myself again :p.

    • one year ago
  13. KonradZuse
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    1/ad-bc [d -b -c a]?

    • one year ago
  14. KonradZuse
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    :P

    • one year ago
  15. tkhunny
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    No, the Determinat CAN be used as part of finding an Inverse, depending on how you find an Inverse. However, in this case, there is no Inverse. The determinant will not help you find it.

    • one year ago
  16. KonradZuse
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    ic, so what happens since there is none?

    • one year ago
  17. KonradZuse
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    I got 0 so that makes sense.

    • one year ago
  18. tkhunny
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    Well, just for practice, can you write the Normal Equations? I'll get you started: \(\sum x_{1} + 3\cdot\sum x_{2} = \sum y\) You write the other two.

    • one year ago
  19. KonradZuse
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    3x1 +9x2= y

    • one year ago
  20. KonradZuse
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    -2x1 - 6x2

    • one year ago
  21. tkhunny
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    Whoops! Using the model \(I = \left(a\cdot x_{1} + b\cdot x_{2} - y\right)^{2}\) Forget that first thing. You did exactly what I told you and that was incorrect. We're trying to FIND the coefficients, not just invent them! Find \(\dfrac{\partial I}{\partial a}\) and \(\dfrac{\partial I}{\partial b}\)

    • one year ago
  22. KonradZuse
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    I'm soo konfused.......

    • one year ago
  23. KonradZuse
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    I am able to solve everything up until the end where it asks for the final part, that's where IU'm confused. I'll show yu what my book explains.

    • one year ago
  24. KonradZuse
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    • one year ago
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  25. Zarkon
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    for your first question just solve the system (without inverses). you will get an infinite number of solutions. which is fine

    • one year ago
  26. KonradZuse
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    yeah I figured that out 2/7-3*t is x1 and x2 is t.

    • one year ago
  27. KonradZuse
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    part b we already solved the normal eq, now I'm working on the error part 3...

    • one year ago
  28. KonradZuse
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    I'm not sure how to get the decimal answer. My matrix comes out as Vector(3, {(1) = 5/7, (2) = 1/7, (3) = 4/7})

    • one year ago
  29. Zarkon
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    are you taking the error to be \[\|Ax-b\|\]

    • one year ago
  30. KonradZuse
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    first it shows the error vector, then shows that.

    • one year ago
  31. KonradZuse
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    is that asking for the norm?

    • one year ago
  32. Zarkon
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    then just compute it...you have all the pieces yes

    • one year ago
  33. KonradZuse
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    does that error play into it, or what exactly am I computing the norm with?

    • one year ago
  34. KonradZuse
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    error vector*

    • one year ago
  35. Zarkon
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    you are computing \(\|Ax-b\|\) where for any vector \(v=<a,b,c>\) \[\|v\|=\sqrt{a^2+b^2+c^2}\]

    • one year ago
  36. KonradZuse
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    SO I guess we do use that error vector then lol.

    • one year ago
  37. KonradZuse
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    (1/7)*sqrt(42)

    • one year ago
  38. KonradZuse
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    0.9258

    • one year ago
  39. Zarkon
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    seems fine

    • one year ago
  40. KonradZuse
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    Thanks for the help :).

    • one year ago
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