# 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?

• Lemma 5.49 in my Notes on the combinatorial fundamentals of algebra ( web.mit.edu/~darij/www/primes2015/sols.pdf ; if numbering changes, see github.com/darijgr/detnotes/releases/tag/2016-09-04 for the version where it is Lemma 5.49). But probably there is a really short answer, depending on how much you know about determinants and permutations. – darij grinberg Oct 4 '16 at 16:46
• Is there a non-combinatorial way of solving this since I am only starting to learn combinatorics.. :/ – Stoner Oct 4 '16 at 17:00

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.

• Hmm.. In that sense why do I have to multiply $\frac{n}{2}$ by $\frac{n-1}{2}$ in the power of ($-1$)? – Stoner Oct 4 '16 at 16:58
• @Stoner The trick is that $n(n-1)/2$ will have the same parity as $n/2$ (if $n$ is even) or $(n-1)/2$ (if $n$ is odd). – egreg Oct 4 '16 at 17:09
• Ahh alright I kinda get the idea but if you have $2k-1$ for $(-1)^{n(n-1)/2}$, then wouldn't equating that to $(-1)^k$ imply that k will be odd since at the second step you will have $(-1)(-1)^k = (-1)^k$? Also, $n = 2k$ is even does not imply that $k$ is odd, for it can also be even. – Stoner Oct 4 '16 at 17:18
• @Stoner Case $n=2k$: then $\frac{n(n-1)}{2}=k(2k-1)$, so $(-1)^{n(n-1)/2}=\bigl((-1)^k\bigr)^{2k-1}$. Not to $(-1)^k\cdot(-1)^{2k-1}$. It's immaterial whether $k$ is even or odd, because $2k-1$ is odd anyway, so raising $(-1)^k$ to $2k-1$ does nothing. – egreg Oct 4 '16 at 17:21
• Oh no my bad.. Sorry about that. Ok I get it now.. So basically for $((-1)^k )^2k-1$, say k is odd. Then the result will still be odd since an odd number multiplied to an odd number is still odd. Similarly, an odd number multiplied to an even number will result in an even number. Hence the power of 2k-1 has "no effect" on the result we seek. Am I right to say that as well? – Stoner Oct 4 '16 at 17:26

$B = \begin{bmatrix} 0&0&\cdots&0&1\\ 0&0&\cdots&1&0\\ \vdots&\cdots&\ddots&\vdots\\ 0&1&\cdots&0&0\\ 1&0&\cdots&0&0\end{bmatrix}A$

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 $n \equiv 1,0 \pmod 4$

• I'm sorry but my understanding of the applications of modulo are still rather fundamental.. How does modulo fit into the context of the powers? – Stoner Oct 4 '16 at 17:08
• We have a pattern $1,-1,-1,1,1,-1,-1\cdots$ How do you want to express it? Modulo 4 is more intuitive to me than $( -1)^{\frac {n^2-n}{2}}$ Maybe the other make more sense to you. – Doug M Oct 4 '16 at 17:57
• I'm sorry but I am still somewhat confused.. What kind of result will we get when we apply n = 2 (mod 4) for example? :( – Stoner Oct 5 '16 at 2:00
• $n=2? -1.$ swap the columns in a $2\times 2$ matrix and the sign of the determinant will flip. – Doug M Oct 5 '16 at 2:47

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 $n\bmod4$.

• Am I correct to view the powers of ($-1$) as the sum of the number of row interchanges? – Stoner Oct 4 '16 at 17:06
• @Stoner - yes indeed. – nbubis Oct 4 '16 at 17:11
• Ah! Then am I also right to say that it will be a summation to $n-1$ i.e sum of x from x = 1 to n-1, which thus results in the formula from the arithmetic progression: n(n-1)/2? – Stoner Oct 4 '16 at 17:21