Entries in every row add to zero (nullspace/determinant question)

[mace2]

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!

 

[HallsofIvy]

What is
\left(\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right)\left(\begin{array}{c}1 \\ 1 \\ 1\end{array}\right)?

What if a+ b+ c= 0, d+ e+ f= 0, g+ h+ i= 0?

 

[mace2]

Thank you HallsofIvy, all so simple now. Just had a little trouble reading it!