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Thread: Finding the matrix T(x)

  1. #1
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    Finding the matrix T(x)



    "Let T be a linear transformation on the plane with

    T \begin{pmatrix}2\\ 3\end{pmatrix} = \begin{pmatrix}-3\\ 3\end{pmatrix} and T \begin{pmatrix}3\\ 4\end{pmatrix} = \begin{pmatrix}5\\ -4\end{pmatrix}


    Solve for the matrix T(x)."


    I tried to insert the various values of T into the matrix and managed to set up some equations, but I am not sure how you would generalize the entire thing in terms of x.

    Any help?

  2. #2
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    I assume you mean:
    [tex]\displaystyle
    \begin{align*}\\
    T\begin{pmatrix}2\\3\end{pmatrix}&=\begin{pmatrix}-3\\3\end{pmatrix}\\
    T\begin{pmatrix}3\\4\end{pmatrix}&=\begin{pmatrix} 5\\-4\end{pmatrix}
    \end{align*}
    [/tex]

    Note: you should surround LaTeX code with [ t e x ]...[/ t e x ] (without the spaces) instead of $$...$$.

    You should try to get the images of the unit vectors, as these correspond to the columns of the matrix.

    For example, to get 0 in the second row, you would use the relation:
    .
    [tex]\displaystyle
    -4\begin{pmatrix}2\\3\end{pmatrix}+3\begin{pmatrix} 3\\4\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}
    [/tex]

    Note that, by luck, the first component is 1; in the general case, you may have to multiply both vectors by a constant to get a unit vector.

    We have therefore:
    [tex]\displaystyle
    \begin{align*}\\
    T\begin{pmatrix}1\\0\end{pmatrix}&= -4\begin{pmatrix}-3\\3\end{pmatrix}+3\begin{pmatrix}5\\-4\end{pmatrix}\\
    &= \begin{pmatrix}27\\-24\end{pmatrix}
    \end{align*}
    [/tex]

    and this gives you the first column of the matrix. You can then use the same technique to get the second column.
    Last edited by barrick; 11-24-2017 at 08:43 AM.

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