Find conditions on $x$ and $y$ which guarantee that one can locally solve the following for $u(x, y)$ and $v(x, y)$

Question

My understanding of this question is that I need to show that the following equations can be solved where $u$ and $v$ can be written as a function of $x$ and $y$.

$xu^2+yv^2=9$
$xv^2-yu^2=7$

I rewrote this as $F(x, y, u, v)$, where the zero locus of $F$ is the set of solutions to the original ewuations:
$F(x, y, u, v) = \begin{pmatrix}xu^2+yv^2-9\\xv^2-yu^2-7\end{pmatrix}$
I took the derivative of $F$ with respect to $u$ and $v$ to get:
$F_{(u, v)} = \begin{pmatrix}2xu&2yv\\-2yu&2xv\end{pmatrix}$
The determinant of this is: $4uv(x^2+y^2)$ but I don't see how conditions on just $x$ and $y$ can guarantee that the determinant is never zero, which is what I need to show in order for $u$ and $v$ to be written as functions of $x$ and $y$.

Answer 1

You have the linear system

$$\begin{pmatrix}x&y\\-y&x\end{pmatrix}\begin{pmatrix}u^2\\v^2\end{pmatrix}=\begin{pmatrix}9\\7\end{pmatrix}.$$

As long as $x$ and $y$ are not both zero, one can solve this in the usual way to get

$$u^2=\frac{9x-7y}{x^2+y^2}\qquad v^2=\frac{7x+9y}{x^2+y^2}.$$

We need both of these to be nonnegative. Thus we need:

• at least one of $x$ and $y$ nonzero
• $9x-7y\geq 0$, and
• $7x+9y\geq 0$.