Correct. w and z would be free variables in this case. So you can define it as anything you want.
You can also think of the matrix like this (where the last two variables, in this case w and z, are free variables):
Hello, I believe I have most of a homogeneous linear system solved. The coefficient matrix is:
That amounts the the following system:
1x +0y + 0w + 0z = 0
0z + 1y + 1w + 0z = 0
x = 0
y + w = 0
I let the parameter t = w and got:
x = 0
y = -t
w = t
z = ?
I think z equals another parameter, perhaps s, but I am not certain.