<s>[INST] A system of equations with degree 2 Let $a,b,c,d \in \mathbb R$. Suppose the following holds
\begin{align} 
a^2-c^2 &=1 \\
b^2-d^2 &=-1 \\
ad-bc &= \pm1 \\
ab-cd &=0
\end{align}
How can I find
$a,b,c$ and $d$. I'm trying to show something regarding differential geometry
and arrive with these equations. I tried to solved it but my algebra sucks.  
Any help or hints on how to solved this system of equations would be much appreciated!
 [/INST] Just two by two matrices,
$$
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{array}
\right)
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) = 
\left(
\begin{array}{cc}
1&0 \\
0&1 \\
\end{array}
\right)
$$
so that the right hand factor must be the inverse of $ M= 
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{array}
\right)$
When $M$ has determinant $1,$  we are demanding
$$
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
d&-c \\
-b&a \\
\end{array}
\right) = M^{-1}
$$
so that $a=d$ and $b=c,$
or
$$
M=
\left(
\begin{array}{cc}
a&c \\
c&a \\
\end{array}
\right)
$$
When $M$ has determinant $-1,$  we are demanding
$$
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
-d&c \\
b&-a \\
\end{array}
\right) = M^{-1}
$$
so that $a=-d$ and $b=-c,$
or
$$
M=
\left(
\begin{array}{cc}
a&c \\
-c&-a \\
\end{array}
\right)
$$
One may fill in both versions using $\cosh t$ and $\sinh t$
</s>