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 
$$\begin{vmatrix}
a & b & c & d\\ 
b & -a & d &-c \\ 
c & -d & -a &b \\ 
d & c & -b &-a 
\end{vmatrix} \neq 0$$
 A: This is a well-known determinant arising from quaternions. It has the value
$$
-(a^2+b^2+c^2+d^2)^2.
$$
Over the real numbers the value can only be zero if all of $a,b,c,d$ are zero.
The quaternion algebra $\mathbb{H}$ can be represented as the set of the above $4\times 4$ matrices with the $4$ real parameters $a,b,c,d$.
The determinant has been already calculated by Dickson in A Matrix Defined by the Quaternion Group, page $245$, in the year $1902$ (three rows differ by a factor of $-1$, so the result there is $+(a^2+b^2+c^2+d^2)^2$).
A: There are several references you might find useful.


*

*Determinant of block matrices. In particular, for $2\times 2$ real matrices $A,B,C,D$, the following formula holds WHEN $CD=DC$,
$$
{\displaystyle \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(AD-BC).}
$$ 
Here is a proof of this formula.

*The matrix representation of quaternions.

[Added] To elaborate Artem's answer, here is another way to do it. 
Denote the matrix in your question as
$$
A=\left(\begin{matrix}
a & b & c & d\\ 
b & -a & d &-c \\ 
c & -d & -a &b \\ 
d & c & -b &-a 
\end{matrix}\right).
$$
Then 
$$
AA^t=
\left(\begin{matrix}
a^2+b^2+c^2+d^2 & 0 & 0 & 0\\ 
0 & a^2+b^2+c^2+d^2 & 0 & 0 \\ 
0 & 0 & a^2+b^2+c^2+d^2 & 0 \\ 
0 & 0 & 0 &a^2+b^2+c^2+d^2 
\end{matrix}\right).
$$
Note that
$$
\det(A)^2=\det(A)\det(A^t).
$$
A: You do not need any references or any remotely complicated calculations to show what is required. The matrix is indeed related to quaternion representation, but it is not the discussed representation.
First, note that if you factor out $-1$ from 3 last rows of your matrix you will get the skew-symmetric matrix
$$
(-1)\begin{vmatrix}
a & b & c & d\\ 
-b & a & -d &c \\ 
-c & d & a &-b \\ 
-d & -c & b &a 
\end{vmatrix}.
$$
Now this is the representation two other answers refer to. Take this matrix and multiply by its transpose. You will find the diagonal matrix with all four elements on the main diagonal $(a^2+b^2+c^2+d^2)$. Now conclude...
A: 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}
