I want to determine the Jacobian matrix $J_f(x,y,z)$ of the image $f: \mathbb{R^3} \to \mathbb{R^3}$ with $f(x,y,z) := (4y, 3x^2-2\sin(yz), 2yz)$.

So we have 3 coordinate functions $f_1,f_2,f_3$ with $$f_1(x,y,z) = 4y$$ $$f_2(x,y,z) = 3x^2 - 2\sin(yz)$$ $$f_3(x,y,z) = 2yz$$

So I get the following matrix for $J_f(x,y,z)$

\begin{pmatrix} 0 & 4 & 0\\ 6x & -2z\cos(zy) & -2\cos(yz)\\ 0 & 2z & 2y \end{pmatrix}

Is that correct or false?

And how can one find out the set of points $a \in \mathbb{R^3}$, for which the Jacobian Matrix $J_f(a)$ is not invertible?


Minor typo when you differentiate $f_2$ with respect to $z$,$$\begin{pmatrix} 0 & 4 & 0\\ 6x & -2z\cos(zy) & -2\color{blue}y\cos(yz)\\ 0 & 2z & 2y \end{pmatrix}$$

It is not invertible when the determinant is zero.

That is when $$4(6x)(2y)=0$$


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