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In order to find the inverse of the function $y = x^3$ where $y = f(x) = x^3$

we need $x = f^{-1}(y)$, which we compute it as $x = y^{\frac{1}{3}}$ so the inverse function.

But how do I calculate the inverse map of the following map? $x \mapsto Ax +By + C$ and $y \mapsto Dx$ ?, where $A,B,C,D$ are real numbers.

I was trying to visualize this in terms of matrices, $\begin{bmatrix}x \\ y\end{bmatrix} \mapsto \begin{bmatrix} Ax + By + C \\ Dx\end{bmatrix}$, may that open up some new insights?

How can we guarantee the existence of the inverse for this two dimensional map?

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  • $\begingroup$ What are $x,y, A, B, C$? Real numbers, integers, matrices? Also, your first statement would seem to indicate $x^2 = y = f(x) = x^3$, was that intended? $\endgroup$
    – qualcuno
    Commented Aug 14, 2018 at 6:03
  • $\begingroup$ Thank you! I just did an edit to correct the typos. $\endgroup$
    – BAYMAX
    Commented Aug 14, 2018 at 6:09
  • $\begingroup$ Great, I'll give this some thought. $\endgroup$
    – qualcuno
    Commented Aug 14, 2018 at 6:10

2 Answers 2

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$x \mapsto Ax +By + C$ and $y \mapsto Dx$ can be rewritten as :

$$\left(\begin{matrix} x' \\ y' \\ 1 \end{matrix} \right)=M\left(\begin{matrix} x \\y\\1\end{matrix} \right)$$

Where

$$M=\left(\begin{matrix} A & B & C \\ D & 0 & 0 \\ 0 & 0 &1\end{matrix} \right).$$

If $M$ is invertible, then $\left(\begin{matrix} x \\y\\1\end{matrix} \right)=M^{-1}\left(\begin{matrix} x' \\ y' \\ 1 \end{matrix} \right)$.

So existence of an inverse is guaranteed by invertibility of the subjacent matrix when you have a linear system.

Which sums up here to $D\ne 0 \ne B$.


Note that we added a row for the affine part. By doing so, we where able to express your function as a linear system. We could have written also :

$$\left(\begin{matrix} x' \\ y' \\ 0 \end{matrix} \right)=M'\left(\begin{matrix} x \\y\\1\end{matrix} \right)$$

But the invertibility would have been more difficult to express in term of matrices.

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  • $\begingroup$ Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A \neq 0$, $D \neq 0$. $\endgroup$
    – qualcuno
    Commented Aug 14, 2018 at 6:18
  • $\begingroup$ Yes indeed. Should I add your remark to the answer ? @GuidoA. $\endgroup$
    – nicomezi
    Commented Aug 14, 2018 at 6:21
  • $\begingroup$ Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark. $\endgroup$
    – qualcuno
    Commented Aug 14, 2018 at 6:29
  • $\begingroup$ Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$ $\endgroup$
    – BAYMAX
    Commented Aug 14, 2018 at 7:04
  • $\begingroup$ Missread, updated. $\endgroup$
    – nicomezi
    Commented Aug 14, 2018 at 7:08
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So you have the mapping $f\colon \mathbb{R}^2\to \mathbb{R}^2$, $f(x,y)=(Ax+By+C,Dx)$. Claim: $f$ has an inverse iff $A,B\not=0$. This follows from the fact that $f$ is invertible iff the linear system of equations $Ax+By=\xi-C, Dx= \eta$ is solvable for all $(\xi,\eta)\in \mathbb{R}^2$. Missprint: The condition is $B,D\not=0$

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  • $\begingroup$ If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective. $\endgroup$
    – qualcuno
    Commented Aug 14, 2018 at 6:09

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