A problem of determinants Suppose that $a,b,c,d \in \mathbb R$. Then
$$\begin{vmatrix} a& -b & -c& -d\\b& a& d& -c\\c& -d& a& b\\d& c& -b& a \end{vmatrix} = 0 \iff a＝b＝c＝d＝0$$  
I can prove this by computing this determinant exactly. But I wonder if there is any brief approach since the matrix has the special structure
$$\begin{bmatrix}A& -B^t\\B& A^t \end{bmatrix}$$
 A: In this case your matrix $M$ satisfies
$$MM^t=\pmatrix{a^2+b^2+c^2+d^2&0&0&0\\
0&a^2+b^2+c^2+d^2&0&0\\
0&0&a^2+b^2+c^2+d^2&0\\
0&0&0&a^2+b^2+c^2+d^2\\
}$$
so that $\det(M)^2=\det(MM^t)=(a^2+b^2+c^2+d^2)^2$. If
$M=\pmatrix{A&-B^t\\B&A^t}$ then
$$MM^t=\pmatrix{AA^t+B^tB&AB^t-B^tA\\BA^t-A^tB&BB^t+A^tA}$$
which will simplify if (as in your example) $A$ and $B^t$ commute.
A: $M=\begin{pmatrix} A & C \\ B & D\end{pmatrix}$ and $A,B,C,D$ commute with each other, then $\det(M)=\det(AD-BC)$
Silvester: Determinant of block matrices 
In our case as Will Jagy noted, it is linked to quaternions, and you can notice that $A=\begin{pmatrix} a & -b\\b & a\end{pmatrix}$ is $(a^2+b^2)\,R(\alpha)$ where $R$ is a rotation matrix.
Thus $B=(c^2+d^2)\, R(\beta)$ and since two rotations in the plane will commute, then $A$ and $B$ are commuting too.
We have also that $A^T=(a^2+b^2)A^{-1}$ since $R^{-1}(\alpha)=R(-\alpha)$.
In particular we deduce that all matrices involved $A,B,A^T,B^T$ commute with each other so 
$\det(M)=\det(AA^T+BB^T)=\det((a^2+b^2)I+(c^2+d^2)I)=(a^2+b^2+c^2+d^2)^2$
