# $f(x, y, z):=\left(x+y^2+z^2, x-y+z, 2 x+y-z\right)$ local and global inverse, calculating $Df^{-1}$

We consider the mapping $$f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$$ given by $$f(x, y, z):=\left(x+y^2+z^2, x-y+z, 2 x+y-z\right) .$$ a) Show that $$f$$ has a local inverse around the point $$(1,-1,2)$$.
b) Calculate $$D f^{-1}(6,4,-1)$$.
c) Can the inverse be defined globally?

a) Do I need to calculate the derivative of the Jacobian matrix and show that it is not equal to zero? Is that enough?
$$D f(1,-1,2)=\left[\begin{array}{ccc} 1 & 2(-1) & 2(2) \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -2 & 4 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{array}\right]$$
$$Det(Df) = 6 \neq 0.$$
b) I can find the inverse of the Jacobian matrix: $$D f(x, y, z)=\left[\begin{array}{lll} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} & \frac{\partial f_1}{\partial z} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial z} \\ \frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y} & \frac{\partial f_3}{\partial z} \end{array}\right]=\left[\begin{array}{ccc} 1 & 2 y & 2 z \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{array}\right]$$
So $$D f(6,4,-1) =\left[\begin{array}{ccc} 1 & 8 & -2 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{array}\right]$$ and I need to find the inverse of the matrix so that I am finished.
c) I don'tt know how to do this.

You are correct for (a). You need to use the inverse function theorem to claim that, since $$Df (1, -1, 2)$$ is invertible, $$f$$ is locally invertible.

For (b), you are calculating $$Df$$, but the question asks you to calculate $$Df^{-1}$$. You seem to think that

$$Df^{-1} (x, y, z) = \text{inverse of } Df (x, y, z),$$

which is not true in general. Instead, one needs to find $$Df^{-1}$$ implicitly using chain rule: since $$f^{-1}$$ is defined by

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

differentiating on both sides at $$(6, 4-1)$$ and using chain rule gives

$$Df (f^{-1}(6, 4, -1) )\circ Df^{-1} (6, 4, -1) = I,$$

since $$f^{-1} (6, 4, -1) = (1, -1, 2)$$ and you have found $$Df(1, -1, 2)$$ in (a), the answer to (b) is the inverse of $$Df(1, -1, 2)$$.

(c) There is no general way to find an inverse of a function (unless it's a linear function). In your particular equation, when one sees terms like $$y^2, z^2$$, one would guess that $$f$$ is not injective (by plugging $$y\mapsto -y$$ or $$z\mapsto -z$$). It almost work if you choose $$y=z$$ (to get rid of $$y-z$$ in the second and third components). Thus one can see that

$$f(x, t, t) = f(x, -t, -t)$$

for all $$t\in \mathbb R$$. Thus $$f$$ is not bijective and does not have a global inverse.

Remark The notation $$f^{-1}$$ is in general reserved for the (global) inverse of $$f$$, while in your question, $$f^{-1}$$ seems to refer to the local inverse of $$f$$ around $$(1, -1, 2)$$, which exists by (a).

• how do you easily know that $f^{-1} (6, 4, -1) = (1, -1, 2)$? Commented May 16, 2023 at 18:54
• I just plug $(1, -1, 2)$ into $f$ and it's $f(1, -1, 2) = (6, 4, -1)$. @Allison Commented May 16, 2023 at 19:07
• Thanks everything is completely clear! I totally understand it. When you have time, could you maybe look at this question math.stackexchange.com/questions/4699861/… as well ? Commented May 16, 2023 at 19:22