Calculating the determinant gives $(a^2+b^2+c^2+d^2)^2$? I need to calculate the following determinant in order to prove the following equality:
$$\det\begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{pmatrix} = (a^2+b^2+c^2+d^2)^2.$$
I tried using Gauss-algorithm to get an easier matrix, but I'm not sure if I did it correctly.
Calling the $4$ lines $I$, $II$, $III$ and $IV$, I did:
(1) $II \cdot a$
(2) $III \cdot a$
(3) $IV \cdot a$
After this I did:
(4) $II' + I \cdot b$
(5) $III' + I \cdot c$
(6) $IV' + I \cdot d$
So finally I got the following matrix:
$$\begin{pmatrix} a & b & c & d \\ 0 & a^2+b^2 & bc-ad & ac+bd \\ 0 & ad+bc & a^2+c^2 & cd-ab \\ 0 & bd-ac & ab-cd & a^2+d^2 \end{pmatrix}.$$
I thought this would make the determinant a bit easier, unfortunately I must have done something wrong. Is multiplication with single lines allowed as I have done it?
Thank you very much.
 A: Notice that the rows are perpendicular to each other and the length of the rows are all the same.If you call the matrix $M$ then 
$\det(M^tM)=\det(M^t)\det(M)=\det(M)^2$ But $$
M^tM=
(a^2+b^2+c^2+d^2)\left(
\begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}
\right)$$
Therefor $\det(M^tM)=(a^2+b^2+c^2+d^2)^4\Rightarrow \det(M)=\pm(a^2+b^2+c^2+d^2)^2$
Set $N=\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}M$ then all the rows of $N$ have length $1$. Further $N$ is completely determined by its first row. Now we have $\det(N)=\pm 1$
. Notice that the determinant gives a smooth  function, $$\det:S^3\to \{-1,1\} $$ By sending $(a,b,c,d)$  to our matrix  and then to the determinant of this matrix. (This is smooth because it is a polynomial function). But the point $(1,0,0,0)\in S^3$ is mapped to $1$. Now we can conclude that because $S^3$ is connected, $\{-1,1\}$ is discrete, and our function is continuous, that the entire $S^3$ is mapped to $1$. Thus $\det(N)=1$ and also $\det(M)=(a^2+b^2+c^2+d^2)^2$
