That's a nice little problem! I get, as you do, , , , and .
Since the value of is not part of the solution, I often find that eliminating it first by dividing one equation by another is helpful.
For example, dividing the first equation, , by the fourth, , gives or which reduces to . That is, either z= x or z= -x.
Similarly, dividing the second equation, , by the third, , gives or which reduces to . We can rewrite that as and so either y= w or y+ w= 1 which gives y= 1- w.
So we have 4 possible combinations:
1) z= x and y= w
2) z= x and y= 1- w
3) z= -x and y= w
4) z= -x and y= 1- w
Putting each of those into the constraint xz- wy= 1 will give solutions. There is, of course, no one matrix that will work- there is a set of matrices with determinant 1 that are equally close to the given matrix.