Finding $\det(B)$ in terms of $\det(A) $ for matrices $A$ and $B$ 
Q: Let the rows of $A \in M_{n \hspace{1mm}\mathbb x \hspace{1mm}n }$($\mathbb F)$ be $a_1,a_2,...,a_n$, and let $B$ be the matrix in which the rows are $a_n,a_{n-1} ,...,a_1$. Calculate $\det(B)$ in terms of $\det(A)$. A: $\det(B) = (-1)^{\frac{n(n-1)}{2}}\det(A)$.

I thought of applying row interchange but in doing so I can't see how I can derive te desired result as given above. Any suggestions as to how I can approach this question?
A: You get $B$ by swapping row $1$ with row $n$, row $2$ with row $n-1$ and so on.
If $n$ is even, you do $n/2$ swaps; if $n$ is odd, you do $(n-1)/2$ swaps.
Now, if $n=2k$ is even,
$$
(-1)^{n/2}=(-1)^k,
\qquad
(-1)^{n(n-1)/2}=((-1)^{k})^{2k-1}=(-1)^k
$$
because $2k-1$ is odd.
Do similarly for the case $n=2k+1$.

There is another way to see it. Start swapping row 1 with row 2; the the new row 2 with row 3 and so on until you get
$$
\begin{bmatrix}
a_2 \\
a_3 \\
\vdots \\
a_n \\
a_1
\end{bmatrix}
$$
This requires $n$ swaps. Now $n-1$ swaps are needed to push $a_2$ just above $a_1$, then $n-2$ to push down $a_3$, and so on. In total
$$
n+(n-1)+(n-2)+\dots+2+1=\frac{n(n-1)}{2}
$$
swaps.
A: Row interchange simply adds a factor of $-1$. How many row interchanges are needed to transform $A$ to $B$? The first interchange gives you:
$$a_n, a_2, \ldots a_{n-1}, a_1$$
The second interchange will switch between the second and second to last, etc.
All in all, you would need $\lfloor n/2 \rfloor$ interchanges. Now, we just need to show that:
$$(-1)^{\lfloor n/2 \rfloor} = (-1)^\frac{n(n-1)}{2}$$
Which can be shown easily by considering the four cases for $n\bmod4$.
A: $B = \begin{bmatrix}
0&0&\cdots&0&1\\
0&0&\cdots&1&0\\
\vdots&\cdots&\ddots&\vdots\\
0&1&\cdots&0&0\\
1&0&\cdots&0&0\end{bmatrix}A$
So, what is the determinant of that matrix.  It is either $1$ or it is $-1$
$\prod_\limits{i=1}^{n} (-1)^{(n+1-i)}$
-1 if $n \equiv 2,3 \pmod 4$
1 if $n \equiv 1,0 \pmod 4$
