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Let us have $A=(a_{ij})$ arbitrary matrix and for $B=(b_{ij})$ matrix we have $b_{ij}=(-1)^{i+j}a_{ij}$.

Prove, that $det A = det B$.

I tried with some examples, and the determinant is same, but how should I prove it in general? I tried to use the general formula(Leibniz formula), but it didn't really seem to help me out. Any ideas?

Thanks!

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    $\begingroup$ I think the best approach is to prove it by induction on the size of the matrix. $\endgroup$ – Daniel Apr 23 '15 at 19:49
  • $\begingroup$ That seems to be an interesting method, thanks. :) $\endgroup$ – Atvin Apr 23 '15 at 19:50
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Let the size of the matrices be $n$. Then $$ \det A = \sum_\sigma \operatorname{sign}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}. $$ Also, $$ \det B = \sum_\sigma \operatorname{sign}(\sigma) \prod_{i=1}^n b_{i, \sigma(i)} = \sum_\sigma \operatorname{sign}(\sigma) \prod_{i=1}^n (-1)^{i+\sigma(i)} a_{i, \sigma(i)}.$$ The point is to now notice that $$\prod_{i=1}^n (-1)^{i+\sigma(i)} = (-1)^{1+2+\dots+n+1+2+\dots+n} = (-1)^{2k} = 1$$ where $k=1+\dots+n$. Hence each term in the sum matches up exactly.

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  • $\begingroup$ Wow, very elegant :) $\endgroup$ – Atvin Apr 23 '15 at 20:00
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Hint. Note that $(-1)^{i+j}a_{ij}=(-1)^ia_{ij}(-1)^j$. Therefore $B=DAD$, where $D=\operatorname{diag}(-1,1,-1,1,\ldots)$.

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