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