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I know I need to use the inverse matrice, but the problem is (the parameter) $k$, because it's a variable that can take any value depending on $k$, but it's not a variable. Think of the derivative $f'(n)(x)$, where $n$ is not a variable. I have no clue on what to do exactly.

$$f_k(m,n)^T=(m-kn,n)^T=\begin{pmatrix}1&-k\\0&1\end{pmatrix}\begin{pmatrix}m\\n\end{pmatrix}$$

What is $f_k^{-1}$?

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    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-10-25
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    wait k doesn't matter right?2012-10-25
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    I just have to find the inverse matrice, right?2012-10-25
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    your matrix (in the picture) is false, it should be $\begin{pmatrix}1&-k\\0&1\end{pmatrix}$ to coincide with the line before2012-10-25
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    What formula of the inverse matrix do you know?2012-10-25
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    I just use the trick where you get rid of the identity matrix.2012-10-25
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    And you can do this for the matrix $\begin{pmatrix}1&-k\\0&1\end{pmatrix}$? Where exactly is your problem?2012-10-25
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    I didn't know I could put k into that matrix and now I realized this. You solved my problem. Thanks.2012-10-25

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