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Thread: [SOLVED] Transofrmation Vector

  1. #1
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    [SOLVED] Transofrmation Vector

    Hi,

    The Question:

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


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

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

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

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

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

    What is $\displaystyle A^{-1}$?
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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 $\displaystyle A$ 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?

    $\displaystyle
    A^{-1} = \left[ {\begin{array}{rr} 1 & { 2} \\ { 2} & 3 \\ \end{array} } \right]
    $?
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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 $\displaystyle A$ matrix should be before the original matrix.
    Are you saying that it should be $\displaystyle
    A\left[ {\begin{array}{r}
    3 \\
    { - 2} \\

    \end{array} } \right] = \left[ {\begin{array}{r}
    1 \\
    0 \\

    \end{array} } \right]\;\& \;A\left[ {\begin{array}{r}
    2 \\
    { - 1} \\

    \end{array} } \right] = \left[ {\begin{array}{r}
    0 \\
    { - 1} \\

    \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 $\displaystyle
    A\left[ {\begin{array}{r}
    3 \\
    { - 2} \\

    \end{array} } \right] = \left[ {\begin{array}{r}
    1 \\
    0 \\

    \end{array} } \right]\;\& \;A\left[ {\begin{array}{r}
    2 \\
    { - 1} \\

    \end{array} } \right] = \left[ {\begin{array}{r}
    0 \\
    { - 1} \\

    \end{array} } \right]?$
    No. Sorry. It should be $\displaystyle A (1,0)^T = (3,-2)^T$ and $\displaystyle 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 $\displaystyle A (1,0)^T = (3,-2)^T$ and $\displaystyle 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 $\displaystyle \left[ {\begin{array}{r}
    1 \\
    0 \\

    \end{array} } \right] \to \left[ {\begin{array}{r}
    3 \\
    { - 2} \\

    \end{array} } \right]\;\& \;\left[ {\begin{array}{r}
    0 \\
    { - 1} \\

    \end{array} } \right] \to \left[ {\begin{array}{r}
    2 \\
    { - 1} \\

    \end{array} } \right]$
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