How to prove that the determinant of the following matrix
$$A = \begin{bmatrix} a&-b&-c&-d\\ b&a&d&-c\\ c&-d&a&b\\ d&c&-b&a \end{bmatrix}$$
is $\det A=(a^2+b^2+c^2+d^2)^2$?
Note that $AA^t=(a^2+b^2+c^2+d^2)I_4$, but we just have to $$|\det A|=(a^2+b^2+c^2+d^2)^2$$
Any hint would be appreciated.
A different approach: If $A,B,C,D$ are square matrices and $CD=DC$, then $$\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(AD-BC).$$ Letting $$A=\begin{pmatrix}a&-b\\b&a\end{pmatrix},B=\begin{pmatrix}-c&-d\\d&-c\end{pmatrix},C=\begin{pmatrix}c&-d\\d&c\end{pmatrix},D=\begin{pmatrix}a&b\\-b&a\end{pmatrix},$$ you can check that $CD=DC$ and thus compute $$\det(AD-BC)=\det\begin{pmatrix}a^2+b^2+c^2+d^2&0\\0&a^2+b^2+c^2+d^2\end{pmatrix}=(a^2+b^2+c^2+d^2)^2.$$