 Solution

With $\eta =\mathrm{diag}(1,-1,-1)$, the defining condition for the $(1+2)$-dimensional Lorentz group isTaking determinants gives $(\det \Lambda {)}^2=1$, hence $\det \Lambda =\pm 1$. The norm of the zeroth column givesso $|{\Lambda }^0{}_0|\ge 1$. The Proper orthochronous Lorentz group is the component with $\det \Lambda =1$ and ${\Lambda }^0{}_0\ge 1$.