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roadjester
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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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
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
I take that back, the numbers are a little different
I have just compute the inverse matrix with an individual method. Now we should follow the second part.
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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.
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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.
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Don't attend to inverse matrix. Just follow the second part. suppose I got it.
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I think Mahmit is being really painstaking in trying to explain this. Good work.
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Very nice question. I was really enjoyed. And I hope you could understand it.
I try to give you the same one.
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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.
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?
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.
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Indeed. very interesting.