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## Main Question or Discussion Point

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!

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!

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