# Thread:

1. ## Jacobian matrix exercise

Hi everyone,

I'm trying to do an exercise about jacobian matrix but I don't really know how to do it can you help me? The exercise says:

$H:R \rightarrow R^3$ is a $C^1$ vectorial field in $R$ and $G:R^2 \rightarrow R^3$ is defined as follow:

$G(x,y) = H(x^2+3y)$

Calculate $JH(6)$ knowing that $JG(0,2) = \begin{pmatrix} 0 & 3 \\ 0 & 0 \\ 0 & -6 \end{pmatrix}$

2. Originally Posted by jollysa87
Hi everyone,

I'm trying to do an exercise about jacobian matrix but I don't really know how to do it can you help me? The exercise says:

$H:R \rightarrow R^3$ is a $C^1$ vectorial field in $R$ and $G:R^2 \rightarrow R^3$ is defined as follow:

$G(x,y) = H(x^2+3y)$

Calculate $JH(6)$ knowing that $JG(0,2) = \begin{pmatrix} 0 & 3 \\ 0 & 0 \\ 0 & -6 \end{pmatrix}$
Define $K:\mathbb{R}^2\to\mathbb{R}$ by $K(x,y) = x^2+3y$. Jacobians respect matrix multiplication (that's just a way of expressing the chain rule), so $JG(0,2) = J(H\circ K)(0,2) = JH(6).JK(0,2)$ (since $K(0,2) = 6$).

3. Thank you for the explanation... So I have that:

$JG(0,2) = JH(6) \cdot JK(0,2) \Rightarrow \begin{pmatrix} 0 & 3 \\ 0 & 0 \\ 0 & -6 \end{pmatrix} = \begin{pmatrix} D_xH_1(6) \\ D_xH_2(6) \\ D_xH_3(6) \end{pmatrix} \cdot \begin{pmatrix} 0 & 3 \end{pmatrix}$

I can see that:

$D_xH1(6)*3 = 3 \Rightarrow D_xH1(6) = 1$
$D_xH2(6)*3 = 0 \Rightarrow D_xH2(6) = 0$
$D_xH3(6)*3 = -6 \Rightarrow D_xH3(6) = -2$

So the solution is:

$JH(6)=\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$

Am I right?

4. Originally Posted by jollysa87
So the solution is:

$JH(6)=\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$

Am I right?
Yes.

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