Proving det(B) = -det(A) theorem - am I on the right track? Theorem
Let $A$ be some $n \times n$ matrix, and $B$ be an $n \times n$ matrix
which is the result of performing a row swap operation on $A$. Column swaps won't be touched on for brevity's sake.
Let $b_{ij}$ be an element of $B$ at row $i$ column $j$. 
Let $C_{ij} = (-1)^{i+j}M_{ij}$ be the cofactor of $b_{ij}$, and $M_{ij}$ be its corresponding minor. Substitute $a_{ij}$ and $A$ for the same definitions of minor/cofactor etc.
Then 
$$\det B = \sum^{n}_{k=1} b_{ik}C_{ik}=-\det A= -\sum^n_{k=1}a_{ik}C_{ik}$$
Given that using row expansion off of any row of $A$ always produce $\det A$. Ditto for $\det B$ and any row of $B$.
Problem
I've been in the process of trying to prove this. Here are my thoughts, so far.
If $i$ is the row that's been interchanged with some other row $l \ne i$, then we can make the connection that:
$$a_{ij} = b_{lj}$$
and
$$a_{lj} = b_{ij}$$
Since $i \ne l$, then $(-1)^{l+j}$ doesn't necessarily equal $(-1)^{i+j}$.
But, if both $i$ and $l$ are even or both $i$ and $l$ are odd, then the sign itself of the cofactor won't change, which means that we can't necessarily factor -1 from either row via the outer sign of the cofactor at least.
Question
I'm not looking for someone to give me the proof; I'd like to solve it myself, but what I'm trying to figure out is whether or not I'm on the right track for finding the relationship; if not, what is a good direction to point me in?

To swap the i and j'th row of A, multiply A on the right side such that $M_{i,j} = 1, M_{j,i} = 1, M_{k,k} = 1$ if $k\ne i$, or $k\ne j$ and $0's$ elsewehere.
$MA = B\\
\det MA = \det M \det A\\
\det M = -1$
