Let D={α1,α2} and E={β1,β2} be two bases for the plane, and suppose that β1=5(α1)-3(α2) and β2=-3(α1) + 2(α2)

- anonymous

- jamiebookeater

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- anonymous

a) express α1 in the form α1=b1β1 + b2β2 and find an expression for α2

- anonymous

seems like you need to find a vector that satisfies D....hmm

- anonymous

are you getting \[\alpha1 = b1\left[ 5(a1) - 3(a2) \right] + b2\left[ -3(a1)+2(a2) \right]\]? Then for \[\alpha2\] it might be similar

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## More answers

- anonymous

the answer for \[\alpha _{1}=2\beta _{1}+3\beta _{2}\]

- anonymous

though I don't know how to get to the answer

- anonymous

\[\alpha _{2}=3\beta _{1}+5\beta _{2}\] I have a difficult time understanding most of basis and S-coordinates

- anonymous

Can anyone help me out?

- anonymous

so is this a set?

- anonymous

yup D and E (which are really S and T) are two sets or basis for the plane

- anonymous

how would you construct the matrix?
so that I can eliminate
\[\alpha _{2}\]

- anonymous

sorry, basis and s-coordinates are still really shaky for me

- anonymous

ok here
its simple...you dont need matrix

- anonymous

ok here
its simple...you dont need matrix

- anonymous

\[\beta_{1}= 5\alpha_{1} - 3\alpha_{2}\] (1)
\[\beta_{2}= -3\alpha_{1} + 2\alpha_{2}\] (2)
if you multiply (1) by 2, and if you multiply (2) by 3
you get
\[\2beta_{1}= 10\alpha_{1} - 6\alpha_{2}\]
\[\3beta_{2}= -9\alpha_{1} + 6\alpha_{2}\]
adding the above two equations, you eliminate the parameter \[\alpha_{2}\]
and it becomes
\[\2beta_{1} + \3beta_{2}= \alpha_{1} \]

- anonymous

oh awesome, ok thanks a lot! I'm guessing for \[\alpha _{2}\] we eliminate (alpha)1

- anonymous

yeah exact same way...sets are usually simple to solve if you know the algebra "tricks"

- anonymous

you dont need a lot to solve them i mean...they seem difficult but it usually is very easy

- anonymous

cool, hey since I have you here would you know how to conclude that A is non-singular given the equation \[In=-1/2[A ^{2}-7A+In]A\]

- anonymous

nvm, i think i got it, A[-1/2A^2+7/2A-1/2IN]=In....where [] equal A^-1?

- anonymous

yeah if you find the correct inverse...
check to be sure that that matrix is infact an inverse

- anonymous

How would you do that? A isn't defined

- anonymous

hmmm....its trivial...you let it be arbitrary...look at the determinant for such arbitrary A

- anonymous

So let A be 2x2 matrix [a,b][c,d] and solve?

- anonymous

you can. if its nonzero then A is non-singular

- anonymous

K thanks, going back to the basis question, if the S coordinates of point P are (-9, 3), what are it's T coordinates?

- anonymous

you just plug them in to your answers from i), sorry

- anonymous

yeah you can use either system used to solve to find the T coord.

- anonymous

this is hard but im here to help

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