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Math Help - Linear Algebra.Linear Transformation.Help

  1. #1
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    Linear Algebra.Linear Transformation.Help

    Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.
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  2. #2
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    Inverse Transformation

    Hello ypatia
    Quote Originally Posted by ypatia View Post
    Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.
    Do you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.

    In the question you have here, the transformation can be written in matrix form as

    \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be  gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}

    So, to find the inverse transformation, you'll need the inverse of the matrix \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}, which is \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}

    To find its effect on (x, y, z), just do the matrix multiplication:

    \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix  } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}

    So there's your answer: x,y,z) \rightarrow (z,y-z,x-y)" alt="T^{-1}x,y,z) \rightarrow (z,y-z,x-y)" />

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello ypatiaDo you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.

    In the question you have here, the transformation can be written in matrix form as

    \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be  gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}

    So, to find the inverse transformation, you'll need the inverse of the matrix \begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}, which is \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}

    To find its effect on (x, y, z), just do the matrix multiplication:

    \begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix  } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}

    So there's your answer: x,y,z) \rightarrow (z,y-z,x-y)" alt="T^{-1}x,y,z) \rightarrow (z,y-z,x-y)" />

    Grandad
    Is this right?? "T is not injective and, as such, it does not have an inverse. For an example of the lack of injectivity, T(1,0,1) -->(2,1,1) and T(2,-1,1) -->(2,1,1)." Because if this is right contradicts to the above solution of T^-1. Thanks in advance
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  4. #4
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    The matrix is <br />
\left( {\begin{array}{*{20}c}<br />
   1 & 1 & 1  \\<br />
   1 & 1 & 0  \\<br />
   0 & 0 & 1  \\<br /> <br />
 \end{array} } \right)<br />
which is singular.
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  5. #5
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    Transformation matrix

    Quote Originally Posted by Plato View Post
    The matrix is <br />
\left( {\begin{array}{*{20}c}<br />
   1 & 1 & 1  \\<br />
   1 & 1 & 0  \\<br />
   0 & 0 & 1  \\<br /> <br />
 \end{array} } \right)<br />
which is singular.
    Sorry, yes. I mapped (x,y,z) onto (x+y+z, x+y, x), of course.

    Grandad
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