Row swap changing sign of determinant I was wondering if someone could help me clarify something regarding the effect of swapping two rows on the sign of the determinant. I know that if $A$ is an $n\times n$ matrix and $B$ is an $n\times n$ matrix obtained from $A$ by swapping two rows, then
$$\det(B)=-\det(A)$$
but I don't know how to prove this.
I have been looking for proofs at internet, and read in both in textbooks and lectures notes that are available that this result is very hard to prove and most approaches rely on induction and so was wondering if there is something wrong with using that $\det(AB)=\det(A)\det(B)$ and then writing $B=EA$ where $E$ is an elementary matrix swapping two rows and using this result to get $\det(B)=\det(E)\det(A)=-\det(A)$ (since showing that $\det(E)=-1$ in this case is not that hard).
Yes, your method would work, and it is probably the most elegant possible.

We can without loss of generality assume that $E$ interchanges the first two rows.
This means that we can write $E$ in block-diagonal form:
$$
\left( \begin{array}[ccccc]
00 &amp; 1 &amp; 0 &amp;\dots &amp; 0 \\
1 &amp; 0 &amp; 0 &amp;\dots &amp; 0 \\
0 &amp; 0 &amp; 1 &amp;\dots &amp; 0 \\
... &amp; ... &amp; ... &amp; 1 &amp; ... \\
0 &amp; 0 &amp; 0 &amp; ... &amp; 1
 \end{array}\right)
$$

Now if you know how to calculate the determinant from the usual Laplace algorithm, starting at the bottom line, you see that the only nonzero terms are...

Also, why can we assume it interchanges the first two lines without loss of generality? (Think of what happens if we change a basis...)
If $~i~$ and $~j~$ are two rows of matrix $~A~$ that are interchanged to give matrix $~A^*~$.

Apply to $~A~$ and $~A^*~$ the general recursive formula twice (two-stage recursion) along the two interchanged rows  $~i~$ and $~j~$.

Using the fact that  $~A~$ and $~A^*~$  are identical when rows  $~i~$ and $~j~$ are deleted (which is the case for the determinants of the resulting matrices after the two-stage recursion) and rearranging the co-factors in the sums, you will find that $~\det A = -\det A^*$.