Put the matrix `[(1,2,-1,0),(-1,1,2,1),(0,3,1,1)]` in reduced row echelon form:
Labeling the top row I, the second row II and the third row III we use permissable transformations; we can add a multiple of one row to another and we can switch any rows, and we can multiply any row by a nonzero constant.
This is in reduced row-echelon form.
There are an infinite number of solutions; they are of the form:
`(x,-1/5x+1/5,3/5x+2/5)` (You can get this by solving `x-5/3z=-2/3` for z and substituting into `y+1/3z=1/3` getting an expression for y in terms of x. Now you have expressions for y and z in terms of x. There are other possibilities.