KonradZuse
  • 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.
Linear Algebra
jamiebookeater
  • jamiebookeater
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KonradZuse
  • KonradZuse
\[A =\left[\begin{matrix}1 & 3 \\ 3 & 9\\ -2 & -6\end{matrix}\right]\] \[B = \left(\begin{matrix}1 \\ 1\\ 0\end{matrix}\right)\]
KonradZuse
  • KonradZuse
@UnkleRhaukus @Zarkon
KonradZuse
  • 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.... :(

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KonradZuse
  • 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?
KonradZuse
  • KonradZuse
nope, it's correct.....
KonradZuse
  • KonradZuse
Tk have you come to save me? :P
tkhunny
  • 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: \(\left(A^{T}A\right)^{-1}\left(A^{T}A\right)x = \left(A^{T}A\right)^{-1}A^{T}b\) \(Ix = \left(A^{T}A\right)^{-1}A^{T}b\)
KonradZuse
  • 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
tkhunny
  • tkhunny
Determinant? It's an inverse.
tkhunny
  • 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?
KonradZuse
  • KonradZuse
I thought taking the inverse is the same as taking the determinant?
KonradZuse
  • KonradZuse
I guess I konfused myself again :p.
KonradZuse
  • KonradZuse
1/ad-bc [d -b -c a]?
KonradZuse
  • KonradZuse
:P
tkhunny
  • 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.
KonradZuse
  • KonradZuse
ic, so what happens since there is none?
KonradZuse
  • KonradZuse
I got 0 so that makes sense.
tkhunny
  • 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.
KonradZuse
  • KonradZuse
3x1 +9x2= y
KonradZuse
  • KonradZuse
-2x1 - 6x2
tkhunny
  • 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 \(\dfrac{\partial I}{\partial b}\)
KonradZuse
  • KonradZuse
I'm soo konfused.......
KonradZuse
  • 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.
KonradZuse
  • KonradZuse
1 Attachment
Zarkon
  • Zarkon
for your first question just solve the system (without inverses). you will get an infinite number of solutions. which is fine
KonradZuse
  • KonradZuse
yeah I figured that out 2/7-3*t is x1 and x2 is t.
KonradZuse
  • KonradZuse
part b we already solved the normal eq, now I'm working on the error part 3...
KonradZuse
  • 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})
Zarkon
  • Zarkon
are you taking the error to be \[\|Ax-b\|\]
KonradZuse
  • KonradZuse
first it shows the error vector, then shows that.
KonradZuse
  • KonradZuse
is that asking for the norm?
Zarkon
  • Zarkon
then just compute it...you have all the pieces yes
KonradZuse
  • KonradZuse
does that error play into it, or what exactly am I computing the norm with?
KonradZuse
  • KonradZuse
error vector*
Zarkon
  • Zarkon
you are computing \(\|Ax-b\|\) where for any vector \(v=\) \[\|v\|=\sqrt{a^2+b^2+c^2}\]
KonradZuse
  • KonradZuse
SO I guess we do use that error vector then lol.
KonradZuse
  • KonradZuse
(1/7)*sqrt(42)
KonradZuse
  • KonradZuse
0.9258
Zarkon
  • Zarkon
seems fine
KonradZuse
  • KonradZuse
Thanks for the help :).

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