Edit: My friend just explained it to me, duh! If the row adds to zero then (row)dot(1,1,...,1) = 0! And it is in the nullspace. Thanks anyway! :) Hi all. I've been learning about determinants and I was reading one of the sample problems in my textbook. I thought I understood the chapter, but I don't understand their answer at all. (Note this is not homework! Just trying to learn.) Their question: "If the entries in every row of A add to zero, solve Ax=0 to prove detA=0. If those entries add to one, show that det(A-I)=0. Does this mean detA=1?" Their answer: "If the entries in every row add to zero, then (1, 1, . . . , 1) is in the nullspace: singular A has det = 0. (The columns add to the zero column so they are linearly dependent.) If every row adds to one, then rows of A − I add to zero (not necessarily detA = 1)." I guess I am confused as to why if the entries in each row add to zero, then (1,1,...,1) is in the nullspace? Any help would be much appreciated! Thanks!