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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
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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KonradZuseBest ResponseYou've already chosen the best response.0
\[A =\left[\begin{matrix}1 & 3 \\ 3 & 9\\ 2 & 6\end{matrix}\right]\] \[B = \left(\begin{matrix}1 \\ 1\\ 0\end{matrix}\right)\]
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
@UnkleRhaukus @Zarkon
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
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

KonradZuseBest ResponseYou've already chosen the best response.0
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

KonradZuseBest ResponseYou've already chosen the best response.0
nope, it's correct.....
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Tk have you come to save me? :P
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
Ax = b Introduce the Transpose \(A^{T}Ax = A^{T}b\) Hope that \(A^{T}A\) is nonsingular 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

KonradZuseBest ResponseYou've already chosen the best response.0
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

tkhunnyBest ResponseYou've already chosen the best response.0
Determinant? It's an inverse.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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

KonradZuseBest ResponseYou've already chosen the best response.0
I thought taking the inverse is the same as taking the determinant?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I guess I konfused myself again :p.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
1/adbc [d b c a]?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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

KonradZuseBest ResponseYou've already chosen the best response.0
ic, so what happens since there is none?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I got 0 so that makes sense.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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

tkhunnyBest ResponseYou've already chosen the best response.0
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

KonradZuseBest ResponseYou've already chosen the best response.0
I'm soo konfused.......
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
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

ZarkonBest ResponseYou've already chosen the best response.1
for your first question just solve the system (without inverses). you will get an infinite number of solutions. which is fine
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
yeah I figured that out 2/73*t is x1 and x2 is t.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
part b we already solved the normal eq, now I'm working on the error part 3...
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
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

ZarkonBest ResponseYou've already chosen the best response.1
are you taking the error to be \[\Axb\\]
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
first it shows the error vector, then shows that.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
is that asking for the norm?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
then just compute it...you have all the pieces yes
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
does that error play into it, or what exactly am I computing the norm with?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
you are computing \(\Axb\\) where for any vector \(v=<a,b,c>\) \[\v\=\sqrt{a^2+b^2+c^2}\]
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
SO I guess we do use that error vector then lol.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Thanks for the help :).
 one year ago
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