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anonymous

  • 5 years ago

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)

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  1. anonymous
    • 5 years ago
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    a) express α1 in the form α1=b1β1 + b2β2 and find an expression for α2

  2. anonymous
    • 5 years ago
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    seems like you need to find a vector that satisfies D....hmm

  3. anonymous
    • 5 years ago
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    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

  4. anonymous
    • 5 years ago
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    the answer for \[\alpha _{1}=2\beta _{1}+3\beta _{2}\]

  5. anonymous
    • 5 years ago
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    though I don't know how to get to the answer

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

  7. anonymous
    • 5 years ago
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    Can anyone help me out?

  8. anonymous
    • 5 years ago
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    so is this a set?

  9. anonymous
    • 5 years ago
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    yup D and E (which are really S and T) are two sets or basis for the plane

  10. anonymous
    • 5 years ago
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    how would you construct the matrix? so that I can eliminate \[\alpha _{2}\]

  11. anonymous
    • 5 years ago
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    sorry, basis and s-coordinates are still really shaky for me

  12. anonymous
    • 5 years ago
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    ok here its simple...you dont need matrix

  13. anonymous
    • 5 years ago
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    ok here its simple...you dont need matrix

  14. anonymous
    • 5 years ago
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    \[\beta_{1}= 5\alpha_{1} - 3\alpha_{2}\] (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 \[\2beta_{1} + \3beta_{2}= \alpha_{1} \]

  15. anonymous
    • 5 years ago
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    oh awesome, ok thanks a lot! I'm guessing for \[\alpha _{2}\] we eliminate (alpha)1

  16. anonymous
    • 5 years ago
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    yeah exact same way...sets are usually simple to solve if you know the algebra "tricks"

  17. anonymous
    • 5 years ago
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    you dont need a lot to solve them i mean...they seem difficult but it usually is very easy

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

  19. anonymous
    • 5 years ago
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    nvm, i think i got it, A[-1/2A^2+7/2A-1/2IN]=In....where [] equal A^-1?

  20. anonymous
    • 5 years ago
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    yeah if you find the correct inverse... check to be sure that that matrix is infact an inverse

  21. anonymous
    • 5 years ago
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    How would you do that? A isn't defined

  22. anonymous
    • 5 years ago
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    hmmm....its trivial...you let it be arbitrary...look at the determinant for such arbitrary A

  23. anonymous
    • 5 years ago
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    So let A be 2x2 matrix [a,b][c,d] and solve?

  24. anonymous
    • 5 years ago
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    you can. if its nonzero then A is non-singular

  25. anonymous
    • 5 years ago
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    K thanks, going back to the basis question, if the S coordinates of point P are (-9, 3), what are it's T coordinates?

  26. anonymous
    • 5 years ago
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    you just plug them in to your answers from i), sorry

  27. anonymous
    • 5 years ago
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    yeah you can use either system used to solve to find the T coord.

  28. anonymous
    • 5 years ago
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    this is hard but im here to help

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