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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.

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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).

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  • $\begingroup$ how do you easily know that $f^{-1} (6, 4, -1) = (1, -1, 2)$? $\endgroup$
    – Allison
    Commented May 16, 2023 at 18:54
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    $\begingroup$ I just plug $(1, -1, 2)$ into $f$ and it's $f(1, -1, 2) = (6, 4, -1)$. @Allison $\endgroup$ Commented May 16, 2023 at 19:07
  • $\begingroup$ 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 ? $\endgroup$
    – Allison
    Commented May 16, 2023 at 19:22

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