If we equate columns $$\hat{T} = \left(\begin{array}{cccccc} \alpha_1& \beta_1 & 0 &\dots & \dots &0 \\ \beta_1 & \alpha_2 & \beta_2 &0 &\dots &0 \\ 0& \beta_2 & \alpha_3 & \beta_3 & \dots &0 \\ \dots& \dots & \dots &\dots &\dots & 0 \\ \dots& & &\beta_{n-2} &\alpha_{n-1}& \beta_{n-1} \\ 0& \dots &\dots &0 &\beta_{n-1} & \alpha_{n} \\ \end{array} \right)$$ we obtain $$\hat{A}\hat{q}_k=\beta_{k-1}\hat{q}_{k-1}+\alpha_k\hat{q}_k+\beta_k\hat{q}_{k+1}.$$