prove determinant of rotation matrix is -1 or 1

Question

Rotation matrix determinant is 1 or -1.How can prove this generally.

Assuming that rotation can be around every axis.

Answer 1

If the rotation is by an angle that is a rational multiple of $\pi$, then a finite power of it is the identity, hence $1=\det I=\det A^m=(\det A)^m$. For the general case, note that $\det$ is continuous and rotations by rational angles are a dense subset.

Answer 2

I assume that you mean ratations in $\mathbb{R}^3$, here. Actually, the determinant is alway $1$. To see why, let $v_1$ be a non-null vector of the axis. Let $v_2$ and $v_3$ be two vectors orthogonal to $v_1$ which are also orthogonal to each other. Then, if $r$ is your rotation, there is some $\theta\in\mathbb R$ such that $r(v_2)=\cos(\theta)v_2+\sin(\theta)v_3$ and that $r(v_3)=-\sin(\theta)v_2+\cos(\theta)v_3$. Therefore, the matrix of $r$ with respect to the basis $(v_1,v_2,v_3)$ is$$\begin{pmatrix}1&0&0\\0&\cos(\theta)&-\sin(\theta)\\0&\sin(\theta)&\cos(\theta)\end{pmatrix},$$whose determinant is $1$.