I'm trying to follow the above illustration by the author but am not quite understanding his proof.
He is saying that by replacing the ith row of A by the jth row of A we end up with two rows that are the same and thus the determinant will equal zero. But that doesn't make sense! Replacement of one row by another row (as shown in elementary matrix operations) will just cause the sign of the determinant to switch, not equal zero!
Maybe there is something I'm just not understanding about this proof?
To illustrate the process, take an example with $n=3$.
Thus, let $$ A=\pmatrix{ -1&2&3\\ 4&-5&6\\ 7&8&-9\\ } $$ First, using $i=j=1$, we get $$a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}=(-1)(-3)+(2)(78)+(3)(67)=360$$ which is the same as the calculation of $\det(A)$, expanding along the first row.
Next, using $i=1$ and $j=2$, we get $$a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=(-1)(42)+(2)(-12)+(3)(22)=0$$ which, as the author observes, is the same as the calculation of $\det(A^*)$, where $$ A^*= \pmatrix{ -1&2&3\\ -1&2&3\\ 7&8&-9\\ } $$ and the expansion is done along the second row.
Of course, we knew in advance that the determinant of $A^*$ would come out to zero, since $A^*$ has two equal rows.