Is there any way to check wheter the determinant of a matrix $A$ with $|\text{det }A|=1$ is positive or negative? Let $A\in\text{GL}(n,\mathbb{R})$ with $|\text{det }A|=1$. Is there any way to check wheter $\text{det }A$ is positive or negative without computing it?
 A: In some cases there is.  Of course you have to know something about $A$.  For example, you happen to know that $A$ is in the same connected component of ${\rm GL}(n,{\mathbb R})$ as some other matrix $B$, then $\det A$ and $\det B$ have the same sign. 
A: A linear transformation with $\det(A)=-1$ is orientation-reversing. So it suffice to check $A$'s action on unit vectors
$$
\langle e_{1},\cdots e_{n}\rangle, e_{i}=\langle \cdots 1,\cdots \rangle
$$
and see if the image under $A$ is oriented as the original basis or not. 
While this may be a "slick"  way on the surface, computationally it is the same thing, for we usually check the orientability of a basis via taking the determinant. But for low dimensional cases this might be an easier approach. I do not think it would work for $n=1000$, for example. 
A: For symmetric matrices, it might be faster to compute the signature of the corresponding quadratic form by symmetrical row-column operations. 
If you don't mind doing something much harder than computing the determinant, you could compute the topological degree of $S^{n-1}\to S^{n-1}$ given by $x\mapsto Ax/\|Ax\|$. For regular $A$, it equals the sign of the determinant. I've written a program for degree computation, but practically it works only up to dimension 5 (up to 10 for very simple examples). Computing that for linear maps make sense just for testing.
