Let be $ A,B\in \mathbb{K}^{n,n} $ arbitrary matrices. Then $$ \det\Bigg(\underbrace{\begin{pmatrix}A&B\\B&A\end{pmatrix}}_{=:L}\Bigg)=\det(A+B)\cdot \det(A-B) $$
My idea: I consider $$ \begin{align}&\det(A+B)\cdot \det(A-B)\\[10pt]&=\det(A+B)\cdot 1\cdot \det(A-B)\cdot 1\\[10pt]&=\Big(\det(A+B)\cdot \det(I_n)\Big)\cdot \Big(\det(A-B)\cdot \det(I_n)\Big)\\[10pt]&=\det\left(\begin{pmatrix}A+B&0\\0&I_n\end{pmatrix}\right)\cdot \det\left(\begin{pmatrix}I_n&0\\0&A-B\end{pmatrix}\right)\\[10pt]&=\det\Bigg(\underbrace{\begin{pmatrix}A+B&0\\0&A-B\end{pmatrix}}_{=:M}\Bigg) \end{align}$$
So far so good. No I tried to find a matrix $X\in \mathbb{K}^{n,n}$ such that $ L=X^{-1}\cdot M\cdot X $ but I couldn't.