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)

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

9 years agoseems like you need to find a vector that satisfies D....hmm

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

9 years agothe answer for \[\alpha _{1}=2\beta _{1}+3\beta _{2}\]

9 years agothough I don't know how to get to the answer

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

9 years agoCan anyone help me out?

9 years agoso is this a set?

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

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

9 years agosorry, basis and s-coordinates are still really shaky for me

9 years agook here its simple...you dont need matrix

9 years agook here its simple...you dont need matrix

9 years ago\[\beta_{1}= 5\alpha_{1} - 3\alpha_{2}\] (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 \[\2beta_{1} + \3beta_{2}= \alpha_{1} \]

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

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

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

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

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

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

9 years agoHow would you do that? A isn't defined

9 years agohmmm....its trivial...you let it be arbitrary...look at the determinant for such arbitrary A

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

9 years agoyou can. if its nonzero then A is non-singular

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

9 years agoyou just plug them in to your answers from i), sorry

9 years agoyeah you can use either system used to solve to find the T coord.

9 years agothis is hard but im here to help

9 years ago