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Math Help - Injective function and it's inverse.

  1. #1
    Super Member Showcase_22's Avatar
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    Injective function and it's inverse.

    Define f:\Re^2 \rightarrow \Re^2 by f(x,y)=(3x+2y,-x+5y). Show that f is injective, and find it's inverse f^{-1}
    It's injective if it is a one-to-one function.

    f(x_1,y_1)=(3x_1+2y_1,-x_1+5y_1)
    f(x_2,y_2)=(3x_2+2y_2,-x_2+5y_2)

    I need to show that x_1=x_2, y_1=y_2 if f(x_1,y_2)=f(x_2,y_2).

    f(x_1,y_2)=f(x_2,y_2)

    (3x_2+2y_2,-x_2+5y_2)=(3x_1+2y_1,-x_1+5y_1)

    Comparing the two sides gives x_1=x_2 and y_1=y_2

    Hence f is injective. I think this proof is right.

    The problem arises when I try and find the inverse:

    x=3x+2y
    y=-x

    and:

    -x+5y=y
    -x=-4y
    y=\frac{x}{4}

    I don't think this accomplishes anything but this clearly isn't the right method.
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  2. #2
    Moo
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    Yop,

    I personnally think you easily get confused if you use the same names for the new variables.

    f(x,y)=(3x+2y,-x+5y) \Leftrightarrow (x,y)=f^{-1}(3x+2y,-x+5y)=f^{-1}(u,v)

    so let u=3x+2y and v=-x+5y

    solve for x and y with respect to u and v. this would give something like :
    \left\{\begin{array}{ll} x=\frac{5u}{17}-\frac{2v}{17} \\ y=\frac{u}{17}+\frac{3v}{17} \end{array} \right.
    (check the calculations, I did it quite quickly and it's not the most important part)

    hence :
    (x,y)=f^{-1}(u,v)=\left(\frac{5u}{17}-\frac{2v}{17} ~,~ \frac{u}{17}+\frac{3v}{17}\right)


    hey, this is the inverse function !
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  3. #3
    Super Member Showcase_22's Avatar
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    Hey thanks Moo! I solved the simultaneous equations myself so I know where it's all coming from and got what you did!!!

    I have been confused once before by a sudden change of variables with the same name. I'll look out for it in the future!
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