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## KonradZuse 3 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

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

2. KonradZuse

@UnkleRhaukus @Zarkon

3. KonradZuse

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

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

nope, it's correct.....

6. KonradZuse

Tk have you come to save me? :P

7. tkhunny

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

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

Determinant? It's an inverse.

10. tkhunny

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

I thought taking the inverse is the same as taking the determinant?

12. KonradZuse

I guess I konfused myself again :p.

13. KonradZuse

1/ad-bc [d -b -c a]?

14. KonradZuse

:P

15. tkhunny

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

ic, so what happens since there is none?

17. KonradZuse

I got 0 so that makes sense.

18. tkhunny

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

3x1 +9x2= y

20. KonradZuse

-2x1 - 6x2

21. tkhunny

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

I'm soo konfused.......

23. KonradZuse

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

25. Zarkon

for your first question just solve the system (without inverses). you will get an infinite number of solutions. which is fine

26. KonradZuse

yeah I figured that out 2/7-3*t is x1 and x2 is t.

27. KonradZuse

part b we already solved the normal eq, now I'm working on the error part 3...

28. KonradZuse

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

are you taking the error to be $\|Ax-b\|$

30. KonradZuse

first it shows the error vector, then shows that.

31. KonradZuse

is that asking for the norm?

32. Zarkon

then just compute it...you have all the pieces yes

33. KonradZuse

does that error play into it, or what exactly am I computing the norm with?

34. KonradZuse

error vector*

35. Zarkon

you are computing $$\|Ax-b\|$$ where for any vector $$v=<a,b,c>$$ $\|v\|=\sqrt{a^2+b^2+c^2}$

36. KonradZuse

SO I guess we do use that error vector then lol.

37. KonradZuse

(1/7)*sqrt(42)

38. KonradZuse

0.9258

39. Zarkon

seems fine

40. KonradZuse

Thanks for the help :).

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