-2R1+1R2->R2 -1R1+1R3->R3 2R2+1R3->R3 To be consistent, since all the elements but the last one on the bottom row are 0, the last one must be 0 also: Or if you prefer c = 5a-2b There will only be a unique solution if a=b=c=0 and that unique solution will be (x,y,z) = (0,0,0) Edwin