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