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Let $C\in M_n(\mathbb{R})$ and let $D$ be the matrix that results when the rows of $C$ are written in reverse order. What is the relationship between $\det(C)$ and $\det(D)$?

What I tried: I was considering writing $D$ as $C $with row $i$ replaced by 1 times row $(n-i)$ added to row i $\forall i \in \mathbb{N}, i\leq n$ and then finding the determinant of the new $D$ and subtracting $\det(C)$, which would then equal $1+\det(C)$ I believe, but I think that this may be incorrect.

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    $\begingroup$ What happens to the determinant in general when you swap rows? Have you tried doing a few small examples? $\endgroup$ – Justine Feb 20 '16 at 18:32
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    $\begingroup$ It's important to remember that a zero determinant happens when one of the rows can be reduced. So, a zero is always zero, regardless of the row order. So you can't have an additive constant here. $\endgroup$ – Kaynex Feb 20 '16 at 18:37
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Hint: Let $J$ be the $n$-by-$n$ matrix with $J_{1\leq i,j\leq n}=\delta_{i+j,n+1}$ (i.e. $J_{ij}=1$ along the antidiagonal and zero otherwise). What is the matrix $JC$, and what does this tell you about $\det{D}$?

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  • $\begingroup$ If one intends the elements of the rows to be reversed instead of the order of the rows, one should consider $CJ$ instead. $\endgroup$ – Semiclassical Feb 20 '16 at 19:22
  • $\begingroup$ how do I prove that det(J) is -1? $\endgroup$ – Raton Feb 20 '16 at 22:49
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    $\begingroup$ @Raton: Well, $\det{J}\neq -1$ in general (check $n=4$ for example). But it's always either $\pm 1$, and you'll also find that the sign is $4$-periodic e.g. same for $n=5$ as $n=9$. I'd suggest a combination of expansion by cofactors and induction. $\endgroup$ – Semiclassical Feb 21 '16 at 4:17

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