# Implicit function theorem intuition behind non-zero jacobian determinant

Implicit function Theorem: In the general implicit function theorem for $$m$$ variables and $$m$$ implicit equations in the form \begin{align} \mathbf F(x_1,x_2,\ldots,x_n, u_1, u_2, \ldots, u_m) = 0 \end{align} where $$\mathbf F=\langle F_1, F_2,...,F_m \rangle$$

I have been introduced to the requirement that the square jacobian matrix for $$\mathbf F(u_1,...,u_n)$$ must be invertible which means the determinant should be non zero. This is apparently analogous to requiring $$\frac{\partial f}{\partial y}\ne0$$ for the $$2D$$ case $$F(x,y)=0$$.

Can someone please explain any intuition behind this requirement?

Consider the warm-up exercise, of formulating the implicit function theorem for linear functions. That is, where $$\mathbf F(x,u)=0$$ can be given by a matrix multiplication formula like $$Ax+Cu=b$$ where $$x$$ in an $$n$$-vector and $$u$$ an $$m$$-vector, with matrices $$A$$ and $$C$$ and fixed vector $$b$$. Linear algebra tells us this equation is always uniquely solveable for $$u$$ given $$x$$ precisely when the $$m\times m$$ matrix $$C$$ is non-singular, that is, has non-vanishing determinant.
In this case $$C$$ is the Jacobian.