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roadjester

  • 3 years ago

Linear Algebra mixed with Calculus Let T: P^2-->R^3 be the linear transformation given by T(p(x)) = <p(-3), p(5), p'(2)>, B={1,x, x^2} the standard basis for P^2 and B' = {e1, e2, e3} be the standard basis for R^3. Use the inverse of the matrix [T]_B,B' to find a quadratic polynomial passing through (-3, 75) and (5, 99) and whose tangent line at x=2 has a slope of 13.

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  1. mahmit2012
    • 3 years ago
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    |dw:1352007689346:dw|

  2. roadjester
    • 3 years ago
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    ok, HUH???

  3. roadjester
    • 3 years ago
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    uhh, wouldn't adding the three elementary matricies, and taking the rref, give you the inverse? I just don't know where to go from there

  4. roadjester
    • 3 years ago
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    ok, no idea what you're writing, but the numbers come out to be the same inverse that I got after doing the Gauss-Jordan Algorithm But a little English would be nice. If we don't interact, I have no clue what you're writing

  5. mahmit2012
    • 3 years ago
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    Now second part !

  6. roadjester
    • 3 years ago
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    I take that back, the numbers are a little different

  7. mahmit2012
    • 3 years ago
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    I have just compute the inverse matrix with an individual method. Now we should follow the second part.

  8. roadjester
    • 3 years ago
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    |dw:1352008344540:dw|

  9. roadjester
    • 3 years ago
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    By adding these three columns: 1 0 0 0 1 0 0 0 1 to my original matrix and row reducing, I get that matrix as my inverse. I'm unfamiliar with your technique.

  10. mahmit2012
    • 3 years ago
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    |dw:1352008523111:dw|

  11. mahmit2012
    • 3 years ago
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    |dw:1352008571359:dw|

  12. roadjester
    • 3 years ago
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    MAHMIT!!!

  13. mahmit2012
    • 3 years ago
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    |dw:1352008680377:dw|

  14. roadjester
    • 3 years ago
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    While you have good intentions, just writing the answer with no explanation does me no good since I not only do not learn anything, but I do not know what you are doing. Could you please slow down to explain? Also, just writing out the solution is against the CoC.

  15. mahmit2012
    • 3 years ago
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    |dw:1352008780107:dw|

  16. mahmit2012
    • 3 years ago
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    Don't attend to inverse matrix. Just follow the second part. suppose I got it.

  17. mahmit2012
    • 3 years ago
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    |dw:1352009050198:dw|

  18. mahmit2012
    • 3 years ago
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    |dw:1352009161620:dw|

  19. mahmit2012
    • 3 years ago
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    Is it clear?

  20. Preetha
    • 3 years ago
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    I think Mahmit is being really painstaking in trying to explain this. Good work.

  21. mahmit2012
    • 3 years ago
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    |dw:1352009295173:dw|

  22. mahmit2012
    • 3 years ago
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    Very nice question. I was really enjoyed. And I hope you could understand it.

  23. roadjester
    • 3 years ago
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    Erm, thanks.

  24. mahmit2012
    • 3 years ago
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    I try to give you the same one.

  25. mahmit2012
    • 3 years ago
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    |dw:1352009879051:dw|

  26. mahmit2012
    • 3 years ago
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    |dw:1352010149750:dw|

  27. mahmit2012
    • 3 years ago
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    I always thought how a software can find the polynomial with passing some points and some particular slope. So I figure it out. Thank you.

  28. roadjester
    • 3 years ago
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    Umm, sure, although don't you think writing an actual program, the language (C++, Java, D, Ruby, Python, etc) doesn't matter, would be easier than doing this by hand every time?

  29. mahmit2012
    • 3 years ago
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    I must find out some polynomial for a control system online. I have information with y, y' y'',... y(n). This problem help me a lot.

  30. mahmit2012
    • 3 years ago
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    |dw:1352010808815:dw|

  31. helder_edwin
    • 3 years ago
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    Indeed. very interesting.

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