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Math Help - linear transformation

  1. #1
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    linear transformation

    Let u = (4, 0) and v = (0, -3)
    T is a linear transformation such that:

    T(u) = (4, 4, 4)
    T(v) = (6, 9, 12)

    a) Find the standard matrix of T

    b) Find T(x, y)


    Only problem I'm having here is that I can't really see the difference between the two questions.
    Here's my solution to part a:
    We need to solve a(4, 0) + b(0, -3) = (x, y)
    This gives us the following:
    4a = x
    -3b = y
    a = x/4
    b = -y/3
    Using the linear property of transforms on a(4, 0) + b(0, -3) = (x, y), we get
    T(x, y) = T[a(4, 0)] + b(0, -3)] = aT(4, 0) + bT(0, -3)
    T(x, y) = a(4, 4, 4) + b(6, 9 ,12) = (x, x, x) + (-2y, -3y, -4y)

    T(x, y) = (x 2y, x 3y, x 4y)

    which is, as far as I can see, gives me the solution to part b)!

    What am I missing here?
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  2. #2
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    Quote Originally Posted by Dr Zoidburg View Post
    Let u = (4, 0) and v = (0, -3)
    T is a linear transformation such that:


    T(x, y) = a(4, 4, 4) + b(6, 9 ,12) = (x, x, x) + (-2y, -3y, -4y)

    T(x, y) = (x – 2y, x – 3y, x – 4y)

    which is, as far as I can see, gives me the solution to part b)!

    What am I missing here?
    T(x,y)=x(1,1,1)+y(-2,-3,-4)

    the two vectors above make up the columns of the matrix transfrom.

    A= \begin{bmatrix}<br />
1 && -2 \\<br />
1 && -3 \\<br />
1 && -4 \\<br />
\end{bmatrix}

    Good luck.
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