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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus @Zarkon

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Since 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.... :(

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0There 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
 2 years ago
Best ResponseYou've already chosen the best response.0nope, it's correct.....

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Tk have you come to save me? :P

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Ax = b Introduce the Transpose \(A^{T}Ax = A^{T}b\) Hope that \(A^{T}A\) is nonsingular and calculate it's inverse: \(\left(A^{T}A\right)^{1}\left(A^{T}A\right)x = \left(A^{T}A\right)^{1}A^{T}b\) \(Ix = \left(A^{T}A\right)^{1}A^{T}b\)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 2 years ago
Best ResponseYou've already chosen the best response.0Determinant? It's an inverse.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Unfortunately, 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
 2 years ago
Best ResponseYou've already chosen the best response.0I thought taking the inverse is the same as taking the determinant?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I guess I konfused myself again :p.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.01/adbc [d b c a]?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0No, 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
 2 years ago
Best ResponseYou've already chosen the best response.0ic, so what happens since there is none?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I got 0 so that makes sense.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Whoops! 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
 2 years ago
Best ResponseYou've already chosen the best response.0I'm soo konfused.......

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

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

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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0part b we already solved the normal eq, now I'm working on the error part 3...

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I'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
 2 years ago
Best ResponseYou've already chosen the best response.1are you taking the error to be \[\Axb\\]

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0first it shows the error vector, then shows that.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0is that asking for the norm?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1then just compute it...you have all the pieces yes

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0does that error play into it, or what exactly am I computing the norm with?

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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0SO I guess we do use that error vector then lol.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks for the help :).
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