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.
 A: 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.
