<s>[INST] 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
$$\begin{bmatrix}A& -B^t\\B& A^t \end{bmatrix}$$
 [/INST] 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.
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