A matrix $\begin{bmatrix}w&x\\y&z\end{bmatrix}\in\mathbb Z^{2\times 2}$ is unimodular if $$|wz-xy|=1$$ holds.



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> Is there a parametrization of such matrices with $|w||z|-xy=1$
> $$w,z<0<\max(y|z|,|w|x)<\frac{|w||z|+xy}2?$$
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Additionally I would prefer to have the constraint
$$\max(|w|,|z|,x,y)\leq2\min(|w|,|z|,x,y)$$ which I think has to hold automatically if a parametrization exists.



On parametrization of a type of unimodular $2\times2$ integral matrix
You put so many restraints on your variables that there are actually no matrices satisfying all the conditions you want at the same time.


In order to simplify things a bit, note that by replacing the matrix
$\begin{bmatrix}w&x\\y&z\end{bmatrix}$ by $\begin{bmatrix}-w&x\\y&-z\end{bmatrix}$ you might as well want to parametrize $x,y,w,z$ such that:


1. $w,z > 0$
2. $wz-xy = 1$
3. $0 < \max(yz,wx)<\frac{wz+xy}2$


Now according to 3 at least one of x,y has to be > 0, and since both are nonzero the equation $xy = wz-1 \geq 0$ implies that both $x,y > 0$.
From combining 2 and 3. it follows that:


4. $wx < \frac{wz+xy}2 = wz - 1/2$
5. $yz < \frac{wz+xy}2 = xy + 1/2$ hence $ yz \leq xy$


Now 4. implies $x<z$ while 5. implies $z\leq x$ these both obviously can't be satisfied at the same time.


It often helps if you have a set of constraints that you want to be satisfied, to try and find an easier looking set of constraints. The only thing I have been trying to arrive at this prove is just try to simplify things as much as possible, until I arrived at a contradiction.


I have the feeling this might have been a better suited question for stack exchange.


