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now my book says to use the formula A^T A = a^T b

and then solve.... I'll post the pratice problem.

how do we get that solution of x1 and x2? Maybe I'm over tihnking it...

Hmm... That's diff from the formula they say...

like how did they get the answers to the question I posted above?

Construct that delightful matrix. I get [3/7,-2/3]^T
Find \(A^{t}\cdot A\), first.

The one I posted, or their example?

I didn't look at the piosted example. Let's go with your original problem statement.

kk

so we have \[\left[\begin{matrix}2 & -2 \\ 1 & 1 \\ 3 & 1\end{matrix}\right]\]

That's "A".

yup now to find A^t

Which we hope is trivial, except for the type-setting. :-)

A^T should be \[\left[\begin{matrix}2 & 1 & 3 \\ -2 & 1 & 1 \end{matrix}\right]\]

Multiply away and you will be pleasantly surprised at the result.

\[\left[\begin{matrix}14 & 0 \\ 0 & 6\end{matrix}\right]\] is what I got.

Perfect. Now the inverse of that.

The book doesn't do that but okay.

Positive and Diagonal - Definitely NonSingular!

\[\left[\begin{matrix}6 & 0 \\ 0 & 14 \end{matrix}\right]\]

Close! Divide by 84.

I forgot the 1/(ad- bc)

Remember that \(A\cdot A^{-1} = I\)

Yes, that's all you left off. Good work.

1/(16 * 4) - (0 * 0)

= 1/64

84

woop sI did 4 not 6 :p

and 16 not 14 lol...

yeah 1/84

so we don't have to take the inverse, do we?

ahahah Yes, \(16*4 \ne 14*6\)

You did already find the inverse. It is 1/84 times the 6,14 diagonal matrix.

yeah but since the book doesn't, do we have to?

hmmm... Konfused... but if you say so ;P.

You started with Ax = b
We added the transpose: \(A^{t}\cdot A\cdot x = A^{t}\cdot b\)

mhm.

\[\left[\begin{matrix}\frac{1}{14} & 0 \\ 0 & \frac{1}{6}\end{matrix}\right]\]

seems like we are just adding it in :P

shall I solve for A^T B?

and solving the first portion yeilds the identity matrix.

I think I'm understanding where this is going... Instead of having to row reduce....

lke a billion times easier :P

I think I've been doing it the hard way :p

\[\left(\begin{matrix}6 \\ -4\end{matrix}\right)\]

A^Tb

that * the inverse = \[\left(\begin{matrix}\frac{3}{7} \\ \frac{-2}{3}\end{matrix}\right)\]

so x1 = 3/7 and x2 = -2/3???

or am I supposed to solve for that now?!

That's it.

This book really SUCKS..... Sometimes it's okay....

Well, good luck with THAT (the book and author, that is)! (laughing)