Partial derivatives and composition with matrices

Question

Given $f$ and $g$ are each functions sending $(0,0)$ to $(0,0)$, if you know that

$$ [Df]_{(0,0)} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \\ \end{pmatrix} $$

and

$$ [D(f \circ g)]_{(0,0)} = \begin{pmatrix} -1 & 4 \\ 2 & 3 \end{pmatrix} $$

What is the derivative of $g$ at $(0,0)$?

I believe the strategy is to replace two elements in the equation below (recognizing that $g$ and $f$ are switched from normal convention in the given problem):

$ [D(g \circ f)]_{a} = [Dg]_{f(a)} \cdot [Df]_{a} $

Via substitution, I get:

$$ \begin{pmatrix} -1 & 4 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} ? & ? \\ ? & ? \end{pmatrix} $$

However, I don't know how to solve the unknown matrix from here.

In a more generalized sense, is there a way to find matrix A when AB = C and we know B and C? Or, is there another way to solve this problem?

I've only so far seen $Ax = b$ used when $x$ is a column vector. I tried to take the inverse of $A$ and multiply it with $C$, but reverse checking my work, that didn't work out.

(This is problem 2 in Robert Ghrist's textbook in multivariable calculus, volume 2, chapter 5.)

Answer 1

By the chain rule,

$$ D(f \circ g)(x) = Df(g(x)) Dg(x) $$

and since $g(0, 0) = (0, 0)$, we know $Df(0, 0)$ from the given. Now we need to find the matrix $U$ such that

$$ \begin{pmatrix} -1 & 4 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} U. $$

But matrix multiplication is just a linear combination of the columns! Denote the columns of $U$ by $u_1, u_2$. Let the matrix on the left be $B$, with columns $b_1, b_2$. Then you just need to solve the two systems

$$\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} u_1 = b_1 $$

and $$\begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} u_2 = b_2$$

which is an easy task. I leave the rest as an exercise.

Answer 2

I've only so far seen Ax=b used when x is a column vector. I tried to take the inverse of A and multiply it with C, but reverse checking my work, that didn't work out.

Why not? You have the correct approach, and finding the inverse is just a matter of following the rule.

$\begin{align}\mathcal Df_{g(0,0)}\, \mathcal Dg_{(0,0)} &=\mathcal D[f\circ g]_{(0,0)} &\qquad& A~B= C \\[2ex] \mathcal Dg_{(0,0)} &= (\mathcal Df_{(0,0)})^{-1}\,\mathcal D[f\circ g]_{(0,0)}&& B= A^{-1}\, C\\&=\begin{bmatrix}2&-1\\1&0\end{bmatrix}^{-1}\begin{bmatrix}-1 &4\\2&3\end{bmatrix}\\&=\dfrac 11\begin{bmatrix}0&1\\-1&2\end{bmatrix}\begin{bmatrix}-1 &4\\2&3\end{bmatrix} &&\left[\begin{smallmatrix}a & b\\c& d\end{smallmatrix}\right]^{-1}=\tfrac{1}{ad-bc}\left[\begin{smallmatrix}~~~d & -b\\-c&~~~a\end{smallmatrix}\right]\\&~~\vdots\end{align}$

The rest is matrix multiplication.