# Given $f: \mathbb{R}^5\to \mathbb{R}^2$, of class $C^1$. Let $a= (1,2,-1,3,0)$; suppose that $f(a) = 0$ and $Df(a)=$

Given $f: \mathbb{R}^5\to \mathbb{R}^2$, of class $C^1$. Let $a= (1,2,-1,3,0)$; suppose that $f(a) = 0$ and $Df(a)=\begin{bmatrix}1 & 3 & 1 & -1 & 2\\ 0 & 0 & 1 & 2 & -4\end{bmatrix}$

(a) Show there is a function $g : B\to \mathbb{R}^2$ of class $C^1$ defined on an open set $B$ of $\mathbb{R}^3$ such that $f(x_1, g_1(x), g_2(x), x_2, x_3)=0$ for $x=(x_1, x_2, x_3)\in B$, and $g(1, 3, 0)=(2, -1)$.

(b) Find $Dg(1, 3, 0)$

(c) Discuss the problem of solving the equation $f(x) = 0$ for an arbitrary pair of the unknowns in terms of the others, near the point $a$.

For (a), I know that $\frac{\partial f}{\partial x}(a)=\begin{bmatrix}1 & -1 & 2\\ 0 & 2 & -4\end{bmatrix}$ and $\frac{\partial f}{\partial y}(a)=\begin{bmatrix}3 & 1\\ 0 & 1\end{bmatrix}$ and as $\det \frac{\partial f}{\partial y}(a)=3\neq 0$, then I can apply the implicit function theorem to find an open $B\subset \mathbb{R}^3$ and a continuous function $g: B\to \mathbb{R}^2$ such that $f(x_1, g_1(x), g_2(x), x_2, x_3)=0$ for $x=(x_1, x_2, x_3)\in B$, and $g(1, 3, 0)=(2, -1)$.

(b) I know that $Dg(x)=-[\frac{\partial f}{\partial y}(x, g(x))]^{-1}\frac{\partial f}{\partial x}(x, g(x))$, with which $Dg(1,3,0)=-[\frac{\partial f}{\partial y}(1,2,-1,3,0)]^{-1}\frac{\partial f}{\partial x}(1,2,-1,3,0)=-\begin{bmatrix}3 & 1\\ 0 & 1\end{bmatrix}^{-1}\begin{bmatrix}1 & -1 & 2\\ 0 & 2 & -4\end{bmatrix}=\begin{bmatrix}-1/3 & 1 & 2\\ 0 & -2 & 4\end{bmatrix}$

(c) What this says is that I can clear $g_1(x)$ and $g_2(x)$ in terms of $x_1, x_1, x_2$ and $x_3$?

Are you all the right arguments? Thank you very much.

• You may stick to the coordinates $x_1, \ldots , x_5$. $x \in B\subset \mathbb{R}^3 \rightarrow x = (x_1,x_4,x_5)$. Near $a$, $f(x_1,\ldots,x_5) = 0$ defines $g: B \subset \mathbb{R}^3\to \mathbb{R}^2$ with $f(x_1,g_1(x_1,x_4,x_5),g_2(x_1,x_4,x_5),x_4,x_5) = 0$ So, near $a$, the equation $f(x_1,\ldots,x_5) = 0$ has a unique solution and it can - at least numerically - be solved for $x_2$ and $x_3$ in terms of $x_1,x_4,x_5$. For (c): If $x_2,x_3$ are given in terms of $x_1,x_4,x_5$ via $g$, then they represent the unique solution of $f(x_1,\ldots,x_5) = 0$ (near $a$). – trancelocation Feb 18 '18 at 7:13

a) It is right but it should be $\frac{\partial f}{\partial(x_2,x_3)}(a)=3$ instead of $\frac{\partial f}{\partial y}(a)=3$ as you wrote.
b) This is ok but the matrix multiplication gives me $$\begin{bmatrix}-\frac{1}{3} & 1 & -2 \\ 0 & -2 & 4\end{bmatrix},$$ which differs from your result in the minus sign of the upper right 2.
c) Compute \begin{equation*} \begin{aligned} & \det \frac{\partial f}{\partial (x_1,x_2)}(\textbf{a}) = 0 & & \det \frac{\partial f}{\partial (x_1,x_3)}(\textbf{a}) = 1 & & \det \frac{\partial f}{\partial (x_1,x_4)}(\textbf{a}) = 2 \\ & \det \frac{\partial f}{\partial (x_1,x_5)}(\textbf{a}) = -4 & & \det \frac{\partial f}{\partial (x_2,x_3)}(\textbf{a}) = 3 & & \det \frac{\partial f}{\partial (x_2,x_4)}(\textbf{a}) = 6 \\ & \det \frac{\partial f}{\partial (x_2,x_5)}(\textbf{a}) = -12 & & \det \frac{\partial f}{\partial (x_3,x_4)}(\textbf{a}) = 3 & & \det \frac{\partial f}{\partial (x_3,x_5)}(\textbf{a}) = -6 \\ & \det \frac{\partial f}{\partial (x_4,x_5)}(\textbf{a}) = 0. & & & & \end{aligned} \end{equation*} Therefore, we do not expect to be able to solve for $x_1$ and $x_2$ in terms of the others near $\textbf{a}$, neither to solve for $x_4$ and $x_5$. For all other pairs, however, we can assure we can solve the equation $f(\mathbf{x})=\mathbf{0}$ in terms of the other three.