 Solution

For $D=\mathrm{diag}(d_1,d_2,d_3)$, the upper entry of $D^{-1}AD$ in positions $i,i+1$ is $b_i{d}_{i+1}/d_i$, and the corresponding lower entry is $c_i{d}_i/d_{i+1}$. They are equal when $(d_{i+1}/d_i{)}^2=c_i/b_i$. All coefficients used here are positive, so chooseDirect matrix multiplication gives the symmetric matrixIts pattern remains tridiagonal. This is positive diagonal symmetrization of a tridiagonal matrix.