Elementary Methods in Number Theory Exercise 1.2.14

Let $a,b,c,d$ be integers such that $ad-bc=1$.For integers $u$ and $v$,define
\begin{equation}
u'=au+bv
\end{equation}
\begin{equation}
v'=cu+dv
\end{equation}
Prove that $(u,v)=(u',v')$.

Proof:
\begin{equation}
u'c=acu+bcv
\end{equation}
\begin{equation}
v'a=acu+adv
\end{equation}
So
\begin{equation}
u'c-v'a=v(bc-ad)
\end{equation}
So
\begin{equation}
v=v'a-u'c
\end{equation}
\begin{equation}
u=du'-bv'
\end{equation}
So
\begin{equation}
(u,v)\geq (u',v')
\end{equation}and
\begin{equation}
(u',v')\geq (u,v)
\end{equation}
So
\begin{equation}
(u,v)=(u',v')
\end{equation}$\Box$

Remark 1:\begin{equation}
\begin{vmatrix}
a&b\\
c&d\\
\end{vmatrix}=1
\end{equation}

\begin{equation}
\begin{pmatrix}
u'\\
v'
\end{pmatrix}=\begin{pmatrix}
a&b\\
c&d\\
\end{pmatrix}\begin{pmatrix}
u\\
v\\
\end{pmatrix}
\end{equation}

I think there is some relation to geometric meaning(Liear transformation).But I can't find it at present,maybe it is related to this post.

2.Maybe it is also related to complex numbers.For

\begin{equation}
(a+bi)(-c+di)=(-ac-bd)+(ad-bc)i
\end{equation}