Show that if $xu - yv = 1$ and $xv + yu = 0$ then $u$ and $v$ have unique solutions iff $x^2 + y^2 \neq 0$. Show that if $xu - yv = 1$ and $xv + yu = 0$ then $u$ and $v$ have unique solutions iff $x^2 + y^2 \neq 0$. **_Proof_** First, consider a re-arrangement of $xu - yv = 1$, namely $u = \frac{1 + yv}{x}$. Then from $xu + yu = 0$ we have $xv + y\big(\frac{1 + yv}{x}\big) = 0$. Therefore, $v = \frac{-y}{x^2 + y^2}$. Hence, $v$ has a unique solution iff $x^2 + y^2 \neq 0$. If I proceeded in a similar way for showing $u$ has a unique solution, would this be an acceptable proof?

The OP's method looks fine.

Here's a geometrical way to do it:

let

$x^2 + y^2 = r^2 > 0; \tag{1}$

note that

$(x, y) \cdot (v, u) = xv + yu = 0 \tag{2}$

implies $(v, u)$ is normal or perpendicular to $(x, y)$. But the vector $(-y, x)$ is also perpendicular to $(x, y)$, and since we are in two dimensions, any vector normal to $(x, y)$ must be collinear with $(-y, x)$; thus

$(v, u) = c(-y, x) \tag{3}$

for some $0 \
e c \in \Bbb R$. Therefore, if we "$\cdot$" each side with $(-y, x)$ we obtain

$1 = -yv + xu = (-y, x) \cdot (v, u) = c(-y, x) \cdot (-y, x) = c(x^2 + y^2) = cr^2, \tag{4}$

whence

$c = \dfrac{1}{r^2}; \tag{5}$

substituting (5) into (3) yields the solution

$(v, u) = (-\dfrac{y}{r^2}, \dfrac{x}{r^2}); \tag{6}$

note this agrees with $v$ as obtained by our OP R.Evet. The joint solution for $u$ and $v$ is easily checked.