 Solution

In the $s=+1,-1$ basis the transfer matrix and its eigenvalues areFor finite positive temperature, $a,b>0$ and ${\lambda }_{+}>|{\lambda }_{-}|$. Consequently $Z={\lambda }_{+}^L+{\lambda }_{-}^L$ and $L^{-1}\log Z\to \log {\lambda }_{+}$ in the thermodynamic limit. There are $2L$ spins, so the Helmholtz free energy per spin isThe factor $1/2$ is essential: one transfer matrix step counts two spins. For completeness, if the undefined $r$ is retained, set $a_r=A_{-1,-1}^{(r)}$, $d_r=A_{r+1,r+1}^{(r)}$ and $b_r=A_{-1,r+1}^{(r)}$. The same thermodynamic limit gives $f_r=-(2\beta {)}^{-1}\log [(a_r+d_r+\sqrt{(a_r-d_r{)}^2+4b_r^2})/2]$ whenever these are two distinct spin values.