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KonradZuse
 3 years 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}));
KonradZuse
 3 years 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}));

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

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0now my book says to use the formula A^T A = a^T b

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0and then solve.... I'll post the pratice problem.

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

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

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm... That's diff from the formula they say...

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

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0IUc... 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?

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0like how did they get the answers to the question I posted above?

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

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0The one I posted, or their example?

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1I didn't look at the piosted example. Let's go with your original problem statement.

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

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Which we hope is trivial, except for the typesetting. :)

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

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Multiply away and you will be pleasantly surprised at the result.

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

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Side 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.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Perfect. Now the inverse of that.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0The book doesn't do that but okay.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Positive and Diagonal  Definitely NonSingular!

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}6 & 0 \\ 0 & 14 \end{matrix}\right]\]

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Fair 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.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0I forgot the 1/(ad bc)

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Remember that \(A\cdot A^{1} = I\)

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, that's all you left off. Good work.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.01/(16 * 4)  (0 * 0)

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0woop sI did 4 not 6 :p

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0and 16 not 14 lol...

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0so we don't have to take the inverse, do we?

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1ahahah Yes, \(16*4 \ne 14*6\)

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1You did already find the inverse. It is 1/84 times the 6,14 diagonal matrix.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0yeah but since the book doesn't, do we have to?

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, 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.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm... Konfused... but if you say so ;P.

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

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Now 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.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}\frac{1}{14} & 0 \\ 0 & \frac{1}{6}\end{matrix}\right]\]

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1This 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.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0seems like we are just adding it in :P

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0shall I solve for A^T B?

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0and solving the first portion yeilds the identity matrix.

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

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

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0I think I'm understanding where this is going... Instead of having to row reduce....

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0lke a billion times easier :P

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0I think I've been doing it the hard way :p

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}6 \\ 4\end{matrix}\right)\]

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Different 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 \(x = (A^{t}\cdot A)^{1}\cdot A^{t}\cdot b\)

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

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

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0so x1 = 3/7 and x2 = 2/3???

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0or am I supposed to solve for that now?!

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Just 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. :)

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

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Like 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.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0This book really SUCKS..... Sometimes it's okay....

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

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Well, good luck with THAT (the book and author, that is)! (laughing)
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