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Let $X,Y$ be $n \times n$ Matrices with $X = x_{i,j}$ and $Y = y_{i,j}$ with $y_{i,j} = (-1)^{i+j}x_{i,j}$. Show that $\det(Y) = \det(X)$.

Because I don't have to calculate a determinant I think the right tool to prove this is the Leibniz formula, but I don't understand how to use it here.

Thanks in advance.

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    Leibniz is certainly a good way to do it. What is $\prod\limits_{i=1}^n \left(-1\right)^{i+\sigma\left(i\right)}$ for a permutation $\sigma$ of $\left\{1,2,\ldots,n\right\}$ ? – 2017-01-25

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Approaching via Leibniz formula:

$\det(X)=\sum\limits_{\sigma\in S_n}\text{sgn}(\sigma)\prod\limits_{i=1}^nx_{i,\sigma(i)}$

$\det(Y)=\sum\limits_{\sigma\in S_n}\text{sgn}(\sigma)\prod\limits_{i=1}^ny_{i,\sigma(i)}=\sum\limits_{\sigma\in S_n}\text{sgn}(\sigma)\prod\limits_{i=1}^n(-1)^{i+\sigma(i)}x_{i,\sigma(i)}$

Now, note that for any $\sigma$ one has $\prod\limits_{i=1}^n(-1)^{i+\sigma(i)}x_{i,\sigma(i)}=\prod\limits_{i=1}^n(-1)^{i+\sigma(i)}\cdot \prod\limits_{i=1}^nx_{i,\sigma(i)}$

Note further that $\prod\limits_{i=1}^n(-1)^{i+\sigma(i)} = (-1)^{\sum\limits_{i=1}^n (i+\sigma(i))}$

Finally, note that $\sum\limits_{i=1}^n (i+\sigma(i))$ is even since it is equal to $2(\sum\limits_{i=1}^n i)$ as $\sigma$ is a permutation.

Thus, $\prod\limits_{i=1}^n(-1)^{i+\sigma(i)}=1$ and so the $(-1)^{i+\sigma(i)}$ in the expression on the far right for $\det(Y)$ can be completely removed. You are left then with the exact same expression for both $\det(X)$ and $\det(Y)$, showing they are equal.

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Hint. $Y=DXD$ where $D=\operatorname{diag}\left(-1,1,-1,1,\ldots,(-1)^n\right)$. Now, what is $\det D$?