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6.4 #2b In Exercise 2 find the least squares solution of the linear equation Ax = b.
b.
> A = (Matrix(3, 2, {(1, 1) = 2, (1, 2) = 2, (2, 1) = 1, (2, 2) = 1, (3, 1) = 3, (3, 2) = 1})); b = (Vector(3, {(1) = 2, (2) = 1, (3) = 1}));
 one year ago
 one year ago
6.4 #2b In Exercise 2 find the least squares solution of the linear equation Ax = b. b. > A = (Matrix(3, 2, {(1, 1) = 2, (1, 2) = 2, (2, 1) = 1, (2, 2) = 1, (3, 1) = 3, (3, 2) = 1})); b = (Vector(3, {(1) = 2, (2) = 1, (3) = 1}));
 one year ago
 one year ago

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

KonradZuseBest ResponseYou've already chosen the best response.0
now my book says to use the formula A^T A = a^T b
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
and then solve.... I'll post the pratice problem.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
You need a most marvelous construct: \(\left(A^{t}\cdot A\right)^{1}\cdot A^{t}\) It is a beautifu thing.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
how do we get that solution of x1 and x2? Maybe I'm over tihnking it...
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Hmm... That's diff from the formula they say...
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
No, it is the same. It's just the "solved" version given the conclusion that \(\left(A^{t}\cdot A\right)\) is NonSingular.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
IUc... so how do I figure out x1 and x2 at the end? I'm just confused on that part... I understand how to do it if it was just 1 matrix, but a matrix * x = another matrix I'm not so sure about?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
like how did they get the answers to the question I posted above?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Construct that delightful matrix. I get [3/7,2/3]^T Find \(A^{t}\cdot A\), first.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
The one I posted, or their example?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
I didn't look at the piosted example. Let's go with your original problem statement.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
so we have \[\left[\begin{matrix}2 & 2 \\ 1 & 1 \\ 3 & 1\end{matrix}\right]\]
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
yup now to find A^t
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Which we hope is trivial, except for the typesetting. :)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
A^T should be \[\left[\begin{matrix}2 & 1 & 3 \\ 2 & 1 & 1 \end{matrix}\right]\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Multiply away and you will be pleasantly surprised at the result.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
\[\left[\begin{matrix}14 & 0 \\ 0 & 6\end{matrix}\right]\] is what I got.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Side note: Be a little more careful with your notation. So far, you have used "A" and "a' interchangeably. Also, both "T" and "t" for transpose. More consistent behavior will eliminate errors.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Perfect. Now the inverse of that.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
The book doesn't do that but okay.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Positive and Diagonal  Definitely NonSingular!
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
\[\left[\begin{matrix}6 & 0 \\ 0 & 14 \end{matrix}\right]\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Fair enough. I took a look at the example. They are setting up a small system of equestions and asking you to solve the system using any method at your disposal. On method of solving such a system is hte Matrix Inverse. That is what we are doing.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I forgot the 1/(ad bc)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Remember that \(A\cdot A^{1} = I\)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Yes, that's all you left off. Good work.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
1/(16 * 4)  (0 * 0)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
woop sI did 4 not 6 :p
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
and 16 not 14 lol...
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
so we don't have to take the inverse, do we?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
ahahah Yes, \(16*4 \ne 14*6\)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
You did already find the inverse. It is 1/84 times the 6,14 diagonal matrix.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
yeah but since the book doesn't, do we have to?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Yes, the book does. When it introduces x1 and x2, that is exactly what it is doing. The book just isn't using matrix methods to solve the system. This strikes me as rather odd, since this is inherently a matrix problem.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
hmmm... Konfused... but if you say so ;P.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
You started with Ax = b We added the transpose: \(A^{t}\cdot A\cdot x = A^{t}\cdot b\)
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Now we have a system to solve. One way to solve this is to find hte inverse of \(A^{t}\cdot A\), if it has an inverse. It does.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
\[\left[\begin{matrix}\frac{1}{14} & 0 \\ 0 & \frac{1}{6}\end{matrix}\right]\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
This gives: \((A^{t}\cdot A)^{1}\cdot (A^{t}\cdot A)\cdot x = (A^{t}\cdot A)^{1}\cdot A^{t}\cdot b\), using the inverse to solve the system.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
seems like we are just adding it in :P
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
shall I solve for A^T B?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
and solving the first portion yeilds the identity matrix.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
The inverse SOLVES the system, just like substitution or elimination or whatever other method you might use to solve the system.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Yes, that is what hte inverse does. The lefthandside is just 'x' when we are done. This symbolizes the solved system.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I think I'm understanding where this is going... Instead of having to row reduce....
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
lke a billion times easier :P
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I think I've been doing it the hard way :p
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
\[\left(\begin{matrix}6 \\ 4\end{matrix}\right)\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Different problems may benefit from different methods. I think you're not quite done with this one. We don't need just \(A^{t}\cdot b\), we need \(x = (A^{t}\cdot A)^{1}\cdot A^{t}\cdot b\)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
that * the inverse = \[\left(\begin{matrix}\frac{3}{7} \\ \frac{2}{3}\end{matrix}\right)\]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Done! VERY good work. This was cuttingedge stuff 40 years ago. That's WAY closer to current research than your basic 300400 year old calculus. :)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
so x1 = 3/7 and x2 = 2/3???
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
or am I supposed to solve for that now?!
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Just for the record, just this Friday, I did a job for my employer using exactly this technique. My "A" was (462 x 4), so I was pretty glad I didn't have to do it by hand. :)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Damn.... Well that was easier than what the book was doing if it came out all nicely like that :) Thanks!
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Like I said, it is a little odd that they would introduce this technique and just bail at the end, requiring you to resort to Algebra II. Maybe the author thinks you're not ready for the Inverse? Odd, indeed.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
This book really SUCKS..... Sometimes it's okay....
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I have another Q that no one really answered if you don't mind taking a look at it.. I just wanted to be sure I did it correctly.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Well, good luck with THAT (the book and author, that is)! (laughing)
 one year ago
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