Mathematics OpenStudy (anonymous):

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) OpenStudy (anonymous):

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

seems like you need to find a vector that satisfies D....hmm OpenStudy (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 OpenStudy (anonymous):

the answer for $\alpha _{1}=2\beta _{1}+3\beta _{2}$ OpenStudy (anonymous):

though I don't know how to get to the answer OpenStudy (anonymous):

$\alpha _{2}=3\beta _{1}+5\beta _{2}$ I have a difficult time understanding most of basis and S-coordinates OpenStudy (anonymous):

Can anyone help me out? OpenStudy (anonymous):

so is this a set? OpenStudy (anonymous):

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

how would you construct the matrix? so that I can eliminate $\alpha _{2}$ OpenStudy (anonymous):

sorry, basis and s-coordinates are still really shaky for me OpenStudy (anonymous):

ok here its simple...you dont need matrix OpenStudy (anonymous):

ok here its simple...you dont need matrix OpenStudy (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 $\2beta_{1} + \3beta_{2}= \alpha_{1}$ OpenStudy (anonymous):

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

yeah exact same way...sets are usually simple to solve if you know the algebra "tricks" OpenStudy (anonymous):

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

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

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

yeah if you find the correct inverse... check to be sure that that matrix is infact an inverse OpenStudy (anonymous):

How would you do that? A isn't defined OpenStudy (anonymous):

hmmm....its trivial...you let it be arbitrary...look at the determinant for such arbitrary A OpenStudy (anonymous):

So let A be 2x2 matrix [a,b][c,d] and solve? OpenStudy (anonymous):

you can. if its nonzero then A is non-singular OpenStudy (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? OpenStudy (anonymous):

you just plug them in to your answers from i), sorry OpenStudy (anonymous):

yeah you can use either system used to solve to find the T coord. OpenStudy (anonymous):

this is hard but im here to help

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