Given a matrix $A=(a_{ij})_{n\times n}$, let $C_{i,j}$ be the cofactor in position $(i,j)$. By the determinant formula, we have
$$\det A=\sum_{i=1}^n a_{i,1}C_{i,1}.$$
What about if we take a different column for the cofactors, that is
$$\sum_{i=1}^n a_{i,1}C_{i,2}$$
Must this evaluate to zero?
On
Yes. Proof by construction: take a matrix $A$. Make another matrix $B$ by copying $A$'s first row to second row (in effect throwing away $A$'s second row).
Now, determinant of $B$ can be equivalently written as sum of pairwise multiplication of $B$'s second row with cofactors of elements of first row (since first and second rows are identical). Also, $\vert B \vert = 0$ since first and second rows are linearly depenedent.
Thus the statement follows.
Yes, the expansion of the cofactor with a different row (or analagously, column) will always produce zero. To see why, consider the cofactor expansion along the $k$th row $$\sum_{j=1}^n a_{kj}C_{kj}.$$ Notice that each of the cofactors $C_{kj}$ has no knowledge of the the entries of the $k$th row. By definition, the cofactors of the $k$th row do not depend on the $k$th row. Therefore we can view the expansion along the entries of some other row, say row $k' \neq k$, given by $$\sum_{j=1}^n a_{k'j}C_{kj},$$ as replacing the entries of the $k$th row by those of the $k'$th row. In effect, the expansion now thinks that row $k$ and row $k'$ of the matrix are equal, and consequently evaluates to zero.