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KonradZuse

  • 2 years ago

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.

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

  2. KonradZuse
    • 2 years ago
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    @UnkleRhaukus @Zarkon

  3. KonradZuse
    • 2 years ago
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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.... :(

  4. KonradZuse
    • 2 years ago
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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?

  5. KonradZuse
    • 2 years ago
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    nope, it's correct.....

  6. KonradZuse
    • 2 years ago
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    Tk have you come to save me? :P

  7. tkhunny
    • 2 years ago
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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: \(\left(A^{T}A\right)^{-1}\left(A^{T}A\right)x = \left(A^{T}A\right)^{-1}A^{T}b\) \(Ix = \left(A^{T}A\right)^{-1}A^{T}b\)

  8. KonradZuse
    • 2 years ago
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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

  9. tkhunny
    • 2 years ago
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    Determinant? It's an inverse.

  10. tkhunny
    • 2 years ago
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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?

  11. KonradZuse
    • 2 years ago
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    I thought taking the inverse is the same as taking the determinant?

  12. KonradZuse
    • 2 years ago
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    I guess I konfused myself again :p.

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

  14. KonradZuse
    • 2 years ago
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    :P

  15. tkhunny
    • 2 years ago
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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.

  16. KonradZuse
    • 2 years ago
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    ic, so what happens since there is none?

  17. KonradZuse
    • 2 years ago
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    I got 0 so that makes sense.

  18. tkhunny
    • 2 years ago
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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.

  19. KonradZuse
    • 2 years ago
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    3x1 +9x2= y

  20. KonradZuse
    • 2 years ago
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    -2x1 - 6x2

  21. tkhunny
    • 2 years ago
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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 \(\dfrac{\partial I}{\partial b}\)

  22. KonradZuse
    • 2 years ago
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    I'm soo konfused.......

  23. KonradZuse
    • 2 years ago
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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.

  24. KonradZuse
    • 2 years ago
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  25. Zarkon
    • 2 years ago
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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

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

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

  28. KonradZuse
    • 2 years ago
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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})

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

  30. KonradZuse
    • 2 years ago
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    first it shows the error vector, then shows that.

  31. KonradZuse
    • 2 years ago
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    is that asking for the norm?

  32. Zarkon
    • 2 years ago
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    then just compute it...you have all the pieces yes

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

  34. KonradZuse
    • 2 years ago
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    error vector*

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

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

  37. KonradZuse
    • 2 years ago
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    (1/7)*sqrt(42)

  38. KonradZuse
    • 2 years ago
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    0.9258

  39. Zarkon
    • 2 years ago
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    seems fine

  40. KonradZuse
    • 2 years ago
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    Thanks for the help :).

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