I have a function $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ defined as:

$$f(x,y) = (3x-y, x-5y)$$

I proved that it's a bijection, now I have to find the inverse function $f^{-1}$.

Because $f$ is a bijection, it has a inverse and this is true:

$$(f^{-1}\circ f)(x,y) = (x,y)$$

$$f^{-1}(3x-y,x-5) = (x,y)$$

I don't know where to go from here. In a one variable function I would do a substitution of the argument of $f^{-1}$ with a variable and express x with that variable, and then just switch places.

I tried to do a substitution like this:

$$3x-y = a$$ $$x-5y = b$$

And then express $x$ and $y$ by $a$ and $b$ , and get this:

$$f^{-1}(x,y) = (\frac{15x-3y}{42}, \frac{x-3y}{14})$$

But I'm not sure if I'm allowed to swap $x$ for $a$, and $y$ for $b$.

Any hint is highly appreciated.

  • $\begingroup$ Note that $f$ is linear, so you could write it as a matrix and then calculate the inverse. $\endgroup$ – PPR Nov 27 '16 at 19:27
  • $\begingroup$ Recall that when finding the inverse of a bijective function of a singular variable of the form $y=f(x)$ you also "swap the variables." $\endgroup$ – John Wayland Bales Nov 27 '16 at 19:30
  • $\begingroup$ @JohnWaylandBales Yes, should I do it here as well? And, if so, what sould I swap? $\endgroup$ – nikol_kok Nov 27 '16 at 19:32
  • $\begingroup$ Your work is absolutely correct as written. You could reduce the first fraction of your final result by a factor of $3$ so that it also had a denominator of $14$. But you swap $x\leftrightarrow a$ and $y\leftrightarrow b$. $\endgroup$ – John Wayland Bales Nov 27 '16 at 19:38
  • $\begingroup$ @JohnWaylandBales So this is one reliable way of doing this? I mean, I can always swap those two when I'm in a situation like this? $\endgroup$ – nikol_kok Nov 27 '16 at 19:41

You have a linear function here, given by the matrix $$ \begin{bmatrix}3&-1\\1&-5 \end{bmatrix} $$ Can you invert the above?

  • $\begingroup$ @qbert Yes, I know how to find the inverse of that matrix, but I don't know what to do with it :( $\endgroup$ – nikol_kok Nov 27 '16 at 19:31
  • $\begingroup$ Write it as your answer. Your linear map is given by the above matrix, the inverse map will be given by the inverse of the above matrix ( with respect to the standard basis) $\endgroup$ – qbert Nov 27 '16 at 19:41
  • $\begingroup$ If you would like to write it in the same form you can apply the inverse matrix to the vector (x,y) $\endgroup$ – qbert Nov 27 '16 at 19:42
  • $\begingroup$ Don't get me wrong, but I'm not sure what you are saying here. Can you point me to some theory or an explanation of that method? $\endgroup$ – nikol_kok Nov 27 '16 at 19:50
  • $\begingroup$ @nikol_kok If I understood that correctly, you actually get $f^{-1}(a, b) = (\frac{5a - b}{14}, \frac{a - 3b}{14})$, and you're wondering whether it's OK to just swap $a$ and $b$ for $x$ and $y$? Yes, that works just fine. $\endgroup$ – Arthur Nov 27 '16 at 19:55

You can split this into two separate functions $u, v:\Bbb R^2\to \Bbb R$ the following way: $$ u(x, y) = 3x-y\\ v(x, y) = x-5y $$ and we have $f(x, y) = (u(x, y), v(x, y))$. What we want is $x$ and $y$ expressed in terms of $u$ and $v$, i.e. solve the above set of equations for $x$ and $y$, so that you get two functions $x(u, v), y(u, v)$. Then $f^{-1}(u, v) = (x(u,v), y(u,v))$.

  • $\begingroup$ What set of equations am I supposed to solve? $\endgroup$ – nikol_kok Nov 27 '16 at 19:47
  • $\begingroup$ @nikol_kok You should solve the equations $$u= 3x-y\\ v= x-5y$$ for $x$ and $y$. This is exactly corresponding to the fact that in order to find the inverse of, say, $g(x) = 5x + 3$, you solve $g = 5x + 3$ for $x$, only in higher dimension. $\endgroup$ – Arthur Nov 27 '16 at 19:49
  • $\begingroup$ In my question I tried a solution like this. Can you check if it's the same/similar? $\endgroup$ – nikol_kok Nov 27 '16 at 19:52

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