 Solution

For a nonidentity element of ${\mathrm{PSL}}_2(\mathbb{R})$ represented by a determinant-one matrix, the classification is elliptic, parabolic or hyperbolic according as the squared trace is less than, equal to or greater than four. This follows either from the eigenvalues or from the discriminant of its fixed-point equation on the boundary of the upper half-plane.

The first matrix has determinant one and trace two, and is not scalar. It is therefore parabolic. Its fixed-point equation isThus its unique fixed point is $z=-2$ on the real boundary, of multiplicity two; infinity is not fixed and there is no fixed point in the upper half-plane. The second matrix has determinant one and trace $-4$, so it is hyperbolic. Its two boundary fixed points are $(-1\pm \sqrt{3})/2$, consistently with this classification.