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KonradZuse

  • one year 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
    • one year 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
    • one year ago
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    @UnkleRhaukus @Zarkon

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

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

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

  10. tkhunny
    • one year 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
    • one year ago
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    I thought taking the inverse is the same as taking the determinant?

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

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

  14. KonradZuse
    • one year ago
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    :P

  15. tkhunny
    • one year 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
    • one year ago
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    ic, so what happens since there is none?

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

  18. tkhunny
    • one year 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
    • one year ago
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    3x1 +9x2= y

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

  21. tkhunny
    • one year 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
    • one year ago
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    I'm soo konfused.......

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

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

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

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

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

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

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

  34. KonradZuse
    • one year ago
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    error vector*

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

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

  38. KonradZuse
    • one year ago
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    0.9258

  39. Zarkon
    • one year ago
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    seems fine

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

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