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 bydividingone 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.