How to show that the determinant of a special $4 \times 4$ matrix is nonzero? Show that if $a,b,c,d \in \mathbb R$ and at least one is different from $0$, then 
$$\begin{vmatrix}
a & b & c & d\\ 
b & -a & d &-c \\ 
c & -d & -a &b \\ 
d & c & -b &-a 
\end{vmatrix} \neq 0$$

I like the Artem methode but this is a direct methode :
We can assume that $a\neq0$ :
\begin{eqnarray}
\begin{vmatrix}
a & b & c & d\\ 
b & -a & d &-c \\ 
c & -d & -a &b \\ 
d & c & -b &-a 
\end{vmatrix} &=&\frac{1}{a}\begin{vmatrix}
a^2 & b & c & d\\ 
ab & -a & d &-c \\ 
ac & -d & -a &b \\ 
ad & c & -b &-a 
\end{vmatrix} \\
&=&  \frac{1}{a}\begin{vmatrix}
a^2 +b^2+c^2+d^2 & b & c & d\\ 
(ab-ab+dc-cd) & -a & d &-c \\ 
(ac-bd-ac+bd) & -d & -a &b \\ 
(ad+cb-bc-ad) & c & -b &-a 
\end{vmatrix}((C_1)+b(C_2)+c(C_3)+d(C_4)\to (C_1) ) \\
&=& \frac{1}{a}\begin{vmatrix}
a^2 +b^2+c^2+d^2 & b & c & d\\ 
0 & -a & d &-c \\ 
0 & -d & -a &b \\ 
0 & c & -b &-a 
\end{vmatrix} \\
&=& \frac{a^2 +b^2+c^2+d^2}{a}\begin{vmatrix}
 -a & d &-c \\ 
 -d & -a &b \\ 
 c & -b &-a 
\end{vmatrix} \\
&=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2)-d(ad-bc)-c(db+ac) \right)\\
&=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2)-ad^2+dbc-cdb-ac^2 \right)\\
&=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2+b^2+c^2) \right)\\
&=& -(a^2+b^2+c^2+d^2)^2
\end{eqnarray}
