Determinant of a $4 \times 4$ matrix 
If $$ A = \begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix} $$ calculate $\det(A)$.


If you calculate
$$AA^t=\begin{pmatrix}a^2+b^2+c^2+d^2&0&0&0\\0&a^2+b^2+c^2+d^2&0&0\\ 0&0&a^2+b^2+c^2+d^2&0\\0&0&0&a^2+b^2+c^2+d^2\end{pmatrix}$$
$$\det(AA^t)=(a^2+b^2+c^2+d^2)^4\Leftrightarrow (\det(A))^2=(a^2+b^2+c^2+d^2)^4$$
The answer is $\det(A)=(a^2+b^2+c^2+d^2)^2$ and not $\det(A)=-(a^2+b^2+c^2+d^2)^2$. Short of calculating it by hand, why is it not negative?

Because the coefficient of $a^4$ (which is the product of the main diagonal) is $+1$ not $-1$.
