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.

 

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

\begin{equation}

(a+bi)(-c+di)=(-ac-bd)+(ad-bc)i

\end{equation}