What's the easiest/fastest way of calculating the determinant of $$\begin{pmatrix}a & b & c & d\\ -b & a & -d & c\\ -c & d & a & -b \\ -d & -c & b & a\end{pmatrix}$$? The result is $(a^2+b^2+c^2+d^2)^2$. This determinant occurs in Zarhin's trick (if $A$ is an Abelian variety, $(A \times A^t)^4$ is principally polarised).
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Write $X$ for the matrix in question. Then $$ X \cdot X^T = (a^2+b^2+c^2+d^2) I. $$ This implies that $$ \det(X) =f(a,b,c,d)= \pm (a^2+b^2+c^2+d^2)^2 . $$ Now as $$ f(a,0,0,0)=a^4 $$ for all $a \in \mathbf R,$ we have that $$ \det(X)=(a^2+b^2+c^2+d^2)^2 $$ for all $a,b,c,d \in \mathbf R.$
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