Question

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.

Answer 1

\[A =\left[\begin{matrix}1 & 3 \\ 3 & 9\\ -2 & -6\end{matrix}\right]\]

\[B = \left(\begin{matrix}1 \\ 1\\ 0\end{matrix}\right)\]

Answer 2

@UnkleRhaukus @Zarkon

Answer 3

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.... :(

Answer 4

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?

Answer 5

nope, it's correct.....

Answer 6

Tk have you come to save me? :P

Answer 7

Ax = b

Introduce the Transpose

\(A^{T}Ax = A^{T}b\)

Hope that \(A^{T}A\) is non-singular and calculate it's inverse:

\(\left(A^{T}A\right)^{-1}\left(A^{T}A\right)x = \left(A^{T}A\right)^{-1}A^{T}b\)

\(Ix = \left(A^{T}A\right)^{-1}A^{T}b\)

Answer 8

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

Answer 9

Determinant? It's an inverse.

Answer 10

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?

Answer 11

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

Answer 12

I guess I konfused myself again :p.

Answer 13

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

Answer 14

:P

Answer 15

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.

Answer 16

ic, so what happens since there is none?

Answer 17

I got 0 so that makes sense.

Answer 18

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.

Answer 19

3x1 +9x2= y

Answer 20

-2x1 - 6x2

Answer 21

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 \(\dfrac{\partial I}{\partial b}\)

Answer 22

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

Answer 23

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.

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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})

Answer 28

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

Answer 29

first it shows the error vector, then shows that.

Answer 30

is that asking for the norm?

Answer 31

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


yes

Answer 32

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

Answer 33

error vector*

Answer 34

you are computing \(\|Ax-b\|\)

where for any vector \(v=<a,b,c>\)

\[\|v\|=\sqrt{a^2+b^2+c^2}\]

Answer 35

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

Answer 36

(1/7)*sqrt(42)

Answer 37

0.9258

Answer 38

seems fine

Answer 39

Thanks for the help :).