So, the problem is this. I can infer that the determinant of a matrix with two identical rows is equal to $0$ because exchanging two rows negates the determinant of the matrix (which is relatively straightforward to prove). Since there is no net-change in the values of the matrix, the only conclusion to draw is that the determinant of the matrix is indeed $0$.
From this, I was curious how to prove that multiplying a row $r$ with a scalar $a$ and adding it to row $k$ (where $r \not= k$, for the sake of argument) will not change the determinant. So, it could be said that a matrix $G$ contains the presumed change to the matrix.
$$\det{G} = \sum_{j=0}^{n}{G_{kj}C_{kj}(G)}$$ $$\det{G} = \sum_{j=0}^{n}{(F_{kj} +aF_{rj})C_{kj}(F)}$$ $$\det{G} = \sum_{j=0}^{n}{(F_{kj} +aF_{rj})C_{kj}(F)}$$ $$\det{G} = \sum_{j=0}^{n}{F_{kj}C_{kj}(F)} +a\sum_{j=0}^{n}{F_{rj}C_{kj}(F)}$$ $$\det{G} = \sum_{j=0}^{n}{F_{kj}C_{kj}(F)} +a\sum_{j=0}^{n}{F_{rj}C_{kj}(F)}$$
Clearly, the lhs of the addition accounts for $\det{F}$, so it's almost there. Somehow, I must prove to myself that the rhs is $0$ to validate this notion in my head.
$$\det{G} = \det{F} +a\sum_{j=0}^{n}{F_{rj}C_{kj}(F)}$$
All the books I've sampled from thus far didn't really feel like validating the rhs as $0$, they just stated that the rhs is simply the determinant of a matrix $F$ where the contents of row $k$ have been replaced with the values of $r$. I can almost visualize that, but can someone provide a little bit more why everyone seems to take it "for granted". Perhaps I don't see it, but the rhs doesn't seem like anything has been exchanged.
How I see it: The fact that $r$ now stands where $k$ stood seems to imply that $r$ is actually taking $k$'s place (or as if it is taking its place) and the matrix now seems to have two rows $r$, therefore the determinant is evaluating to $0$ and yielding:
$$\det{G} = \det{F}$$
Am I close, too far off? If someone could explain a bit further, I'd be most grateful.
