Questions on the proof of "If you switch two rows on a matrix, then $\det(B) = - \det(A)$" 
Applying to a square matrix $A$ the row operation $R_i \leftrightarrow R_j, i \neq j$ (that is, swapping the $i$-th and $j$-th rows), we obtain a new matrix $B$. Prove that $\det(B) = - \det(A)$.

This is the proof:
$$\det(B) = \sum_{\sigma \in S_n} (-1)^{\sigma}a_{1\sigma(1)} \times \cdots \times a_{j\sigma(i)} \times \cdots a_{i\sigma(j)} \times \cdots \times a_{n \sigma(n)} $$
$$  = \sum_{\sigma \in S_n} (-1)^{\sigma} a_{1 \sigma \tau(1)} \times \cdots \times a_{n \sigma \tau(n)}$$
where $\pmatrix{i & j} = \tau$. As $\sigma$ runs through $S_n$, $\mu = \sigma \tau$ also runs through $S_n$, so we get
$$\sum (-1)^{\tau} a_{1 \mu(1)} \times \cdots \times a_{n \mu(n)} = \sum (-1)^{\tau} (-1)^{\mu} a_{1 \mu(1)} \times \cdots \times a_{n \mu(n)}$$
$\mu = \sigma \tau \implies \mu \tau = \sigma \tau^2 = \sigma \implies$
$$ = - \det(A).$$
I understand the first line, thats fairly straight forward. I start getting confused on the second line:


*

*How does the $\tau$ affect it like it does and so we don't consider the switch in the rows now? 

*In the third line of equations, why do we have $-1$ to the power of $\tau$ but everything else is in terms of $\mu$ and then after the equals sign, why is a random $(-1)^{\mu}$ added and then why does that imply $-\det(A)$?


Can someone answer these questions and explain this proof to me please?
 A: For the first question, note that $\{\sigma:\sigma\in S_n\}=\{\sigma\tau:\sigma\in S_n\}$.
Formula $\sum (-1)^{\tau} a_{1 \mu(1)} \times \cdots \times a_{n \mu(n)}$ si wrong and corrected by $\sum (-1)^{\mu\tau} a_{1 \mu(1)} \times \cdots \times a_{n \mu(n)}$ because the substitution $\sigma=\mu\tau$. The sign of a permutation is a group homomorphism, that's $(-1)^{\mu\tau}=(-1)^\mu(-1)^\tau$. Finally, note that $(-1)^\tau=-1$ because $\tau$ is odd a permutation.
A: The determinant is  is an alternating multilinear form which have the additional property of changing their sign under exchange of two arguments:
$$\det(\cdots,x_i,\cdots,x_j,\cdots)=-\det(\cdots,x_j,\cdots,x_i,\cdots)$$
A: Another proof might be the following. You know the determinant vanishes when two rows are equal. Thus
$$\begin{align}0&=\det(R_1\mid \cdots \mid R_i+  R_j\mid \cdots\mid R_i+  R_j\mid \cdots \mid R_n)\\{}\\&= \det(R_1\mid \cdots \mid R_i+  R_j\mid \cdots\mid   R_j\mid \cdots \mid R_n)\\&+\det(R_1\mid \cdots \mid R_i+  R_j\mid \cdots\mid R_i\mid \cdots \mid R_b)\\{}\\&= \det(R_1\mid \cdots \mid   R_j\mid \cdots\mid   R_j\mid \cdots \mid R_n)\\&+\det(R_1\mid \cdots \mid   R_j\mid \cdots\mid R_i\mid \cdots \mid R_n)\\&+\det(R_1\mid \cdots \mid R_i\mid \cdots\mid R_i\mid \cdots \mid R_n)\\&+\det(R_1\mid \cdots \mid R_i\mid \cdots\mid R_j\mid \cdots \mid R_n)\end{align}$$
In the above we used the determinant is row-wise linear a few times. Now we eliminate the determinants of the two matrices that have equal rows, so we're left with
$$\begin{align}0&=\det(R_1\mid \cdots \mid   R_j\mid \cdots\mid R_i\mid \cdots \mid R_n)\\&+\det(R_1\mid \cdots \mid R_i\mid \cdots\mid R_j\mid \cdots \mid R_n)\end{align}$$
which is precisely saying that 
$$\det(R_1\mid \cdots \mid   R_j\mid \cdots\mid R_i\mid \cdots \mid R_n)=-\det(R_1\mid \cdots \mid R_i\mid \cdots\mid R_j\mid \cdots \mid R_n)$$
