An implicit function theorem question In my course on multivariate calculus we treat the implicit function theorem and I am stuck on the following question: 
Find the values of $a$ and $b$ such that, in a neighbourhood of $(x,y,u,v) = (0,1,1,-1)$, $x$ and $y$ are implicitly defined as $C^1$-functions $f$ and $g$ of $u$ and $v$ by the system of equations :
$$\begin{cases}
e^x +uy + u^4v-a=0 \\
y\cos(x) +bx +b^2u -2v = 4
\end{cases}$$
NB: check whether all conditions of the implicit function theorem are satisfied and state the conclusions as accurately as possible.
My work so far:
Let $G : \mathbb{R}^4 \to \mathbb{R}^2$ be defined by:
$$G(x,y,u,v)=
\begin{pmatrix}
e^x +uy + u^4v -a\\
y\cos(x) +bx +b^2u -2v
\end{pmatrix}
$$
Obviously, this function is $C^1$. 
After this, I presume I have to form a matrix of partial derivatives and check for what values of $a$ and $b$ that matrix is non-singular. 
Could anyone give me some guidance on how to continue?
Thanks in advance.
 A: Well, answer sheet came online so I'll give the answer.
The Jacobian Matrix of the function $G: \mathbb{R}^4 \to \mathbb{R}^2$ defined by:
$$G(x,y,u,v)=\begin{pmatrix}
e^x + uy + u^4 \\
y\cos(x) + bx + b^2u-2
\end{pmatrix}$$
is 
$$DG(x,y,u,v)=\begin{pmatrix}
e^x & u & y+4u^3v & u^4 \\
-y\sin(x) + b & \cos(x) &b^2 & -2
\end{pmatrix}.$$
Because all entries are continuous functions, $G$ is a $C^1$ function.
Observe that $$G(0,1,1,-1) = \begin{pmatrix} 1 \\ b^2+3 \end{pmatrix} = \begin{pmatrix} a \\ 4 \end{pmatrix} \iff a = 1 \wedge b^2=1 \Longrightarrow a = 1 \wedge \left( b=-1 \vee b = 1\right).$$ The matrix with partial derivatives of $G$ with respect to $x$ and $y$ at  $(0,1,1,-1)$, which is
$$\begin{pmatrix} \frac{\partial G}{\partial x}(0,1,1,-1) & \frac{\partial G}{\partial y}(0,1,1,-1) \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ b & 1 \end{pmatrix} $$ is singular if $b=1$ and non-singular if $b=-1$. So for $a=1$ and $b=-1$ all conditions of the implicit function theorem are satisfied and thus there exists an open neighbourhood $B$ of $(1,-1)$, an open neighbourhood $U$ of $(0,1,1,-1)$ and $C^1$ functions $f:B \to \mathbb{R}$ and $g : B \to \mathbb{R}$ such that
$$\{ (f(u,v), g(u,v), u, v : (u,v) \in B \} = \{(x,y,u,v) \in U: G(,x,y,u,v) = (1.4) \}$$ 
A: That's right.  Let $z=[u,v]'$.  Then the implicit function theorem implies in your case that
$$\frac{\partial G}{\partial z'} + \begin{bmatrix} \frac{\partial G}{\partial x} \; \frac{\partial G}{\partial y}\end{bmatrix}
\begin{bmatrix} \frac{\partial f}{\partial z'} \\ \frac{\partial g}{\partial z'} \end{bmatrix}=0.$$
So if the matrix of partial derivatives (with respect to $x,y$) is invertible then life is wonderful.
