Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 3 / 1D (Groups)

Use column vectors in a right-handed frame, and interpret clockwise as viewed from the positive $x$-axis towards the origin. The rotation matrix fixes $e_x$ and sends $e_y$ towards $-e_z$. With $q=1/\sqrt{2}$, its images of the basis vectors giveThe transverse block is the planar rotation through $-\pi /4$. Viewing from the other end of the axis reverses the meaning of clockwise and replaces $R$ by $R^T$; the viewpoint is not specified in the paper.

The orthogonal reflection in $x=y$ exchanges the first two coordinates and leaves the third unchanged. Its fixed subspace is the plane, while its normal $(1,-1,0)$ changes sign. ThereforeThis reflection matrix has determinant $-1$, as expected.

For column vectors, the second transformation multiplies on the left of the first. Thus the required matrix is $SR$, not $RS$. Using the rotation matrix convention in part (i),Indeed, a vector first becomes $(x,qy+qz,-qy+qz)$, and the orthogonal reflection then exchanges its first two entries.