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Math Help - [SOLVED] Transofrmation Vector

  1. #1
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    [SOLVED] Transofrmation Vector

    Hi,

    The Question:

    A transformation x \rightarrow Ax sends vector (1,0)^T to (3,-2)^T and vector (0,-1)^T to (2,-1)^T. What is A? Find the inverse of A.


    This is what I thought you're supposed to do:

    Let A = (x,y)^T. Therefore:

    (1,2)^T (x,y)^T = (3,-2)^T
    (0,-1)^T (x,y)^T = (2,-1)^T

    I thought I could solve for x, y simultaneously but I really don't know where to go .

    Thanks
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  2. #2
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    Does this work? A = \left[ {\begin{array}{rr} 3 & { - 2}  \\   { - 2} & 1  \\ \end{array} } \right] WHY?

    What is A^{-1}?
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  3. #3
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    Yes, it does work. But I'm not too sure why, sorry .

    Oh and I realise that what I wrote was wrong: the A matrix should be before the original matrix.

    I can see that if you know that it is a 2x2 matrix then you can figure it out by normal matrix multiplication.

    Is the transformation matrix always a 2x2 matrix? Why is that?

    <br />
A^{-1} = \left[ {\begin{array}{rr} 1 & { 2} \\ { 2} & 3 \\ \end{array} } \right]<br />
?
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  4. #4
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    Quote Originally Posted by Solo View Post
    Oh and I realise that what I wrote was wrong: the A matrix should be before the original matrix.
    Are you saying that it should be <br />
A\left[ {\begin{array}{r}<br />
   3  \\<br />
   { - 2}  \\<br /> <br />
 \end{array} } \right] = \left[ {\begin{array}{r}<br />
   1  \\<br />
   0  \\<br /> <br />
 \end{array} } \right]\;\& \;A\left[ {\begin{array}{r}<br />
   2  \\<br />
   { - 1}  \\<br /> <br />
 \end{array} } \right] = \left[ {\begin{array}{r}<br />
   0  \\<br />
   { - 1}  \\<br /> <br />
 \end{array} } \right]?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Are you saying that it should be <br />
A\left[ {\begin{array}{r}<br />
   3  \\<br />
   { - 2}  \\<br /> <br />
 \end{array} } \right] = \left[ {\begin{array}{r}<br />
   1  \\<br />
   0  \\<br /> <br />
 \end{array} } \right]\;\& \;A\left[ {\begin{array}{r}<br />
   2  \\<br />
   { - 1}  \\<br /> <br />
 \end{array} } \right] = \left[ {\begin{array}{r}<br />
   0  \\<br />
   { - 1}  \\<br /> <br />
 \end{array} } \right]?
    No. Sorry. It should be  A (1,0)^T = (3,-2)^T and  A (0,-1)^T = (2,-1)^T

    Edit: How did you know it would be a 2x2 matrix?
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  6. #6
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    Quote Originally Posted by Solo View Post
    No. Sorry. It should be  A (1,0)^T = (3,-2)^T and  A (0,-1)^T = (2,-1)^T
    Edit: How did you know it would be a 2x2 matrix?
    O.K. The matrix I gave you in my O.P. is correct.
    It is the transformation that \left[ {\begin{array}{r}<br />
   1  \\<br />
   0  \\<br /> <br />
 \end{array} } \right] \to \left[ {\begin{array}{r}<br />
   3  \\<br />
   { - 2}  \\<br /> <br />
 \end{array} } \right]\;\& \;\left[ {\begin{array}{r}<br />
   0  \\<br />
   { - 1}  \\<br /> <br />
 \end{array} } \right] \to \left[ {\begin{array}{r}<br />
   2  \\<br />
   { - 1}  \\<br /> <br />
 \end{array} } \right]
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