<s>[INST] Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$ 
Let $a, b, c, d \in \mathbb K$ where $\mathbb K$ is a field. Prove that
$$\det \begin{bmatrix}
a & -b & -c & -d\\ 
b & a & -d & c\\ 
c & d & a & -b\\ 
d & -c & b & a
\end{bmatrix} = (a^2+b^2+c^2+d^2)^2$$

I'm looking for a smart way to solve this problem. If we denote 
$$A = \begin{bmatrix}
a & -b \\ 
b & a \\ 
\end{bmatrix}$$ 
and 
$$B = \begin{bmatrix}
-c & -d \\ 
-d & c \\ 
\end{bmatrix}$$
we have that
$$ \begin{bmatrix}
a & -b & -c & -d\\ 
b & a & -d & c\\ 
c & d & a & -b\\ 
d & -c & b & a
\end{bmatrix} = \begin{bmatrix}
A & B \\ 
-B & A \\ 
\end{bmatrix} $$
So it's sufficient to proof that 
$$ \det \begin{bmatrix}
A & B \\ 
-B & A \\ 
\end{bmatrix} = (\det A - \det B)^2. $$
Help?
 [/INST] Generally， if $B$ is symmetric such that $A^TB=BA$, then
$$ \det \begin{bmatrix}
A & B \\ 
-B & A \\ 
\end{bmatrix} = \det (AA^T +BB^T). $$
In fact
$$ \begin{bmatrix}
A & B \\ 
-B & A \\ 
\end{bmatrix}\begin{bmatrix}
A & B \\ 
-B & A \\ 
\end{bmatrix}^T=\begin{bmatrix}
A & B \\ 
-B & A \\ 
\end{bmatrix}\begin{bmatrix}
A^T & -B^T \\ 
B^T & A^T \\ 
\end{bmatrix} = \begin{bmatrix}
AA^T+BB^T & -AB^T+BA^T \\ 
-BA^T+AB^T & AA^T+BB^T \\ 
\end{bmatrix}=\begin{bmatrix}
AA^T+BB^T & 0 \\ 
0 & AA^T+BB^T \\ 
\end{bmatrix} $$
and hence
$$ \det\begin{bmatrix}
A & B \\ 
-B & A \\ 
\end{bmatrix}=\sqrt{\det\begin{bmatrix}
AA^T+BB^T & 0 \\ 
0 & AA^T+BB^T \\ 
\end{bmatrix}}=\det(AA^T+BB^T).$$
</s>