The product of the matrices is \[\begin{pmatrix} 2 & a \\ -3 & -1 \end{pmatrix} \begin{pmatrix} -\frac{1}{16} & b \\ \frac{3}{16} & \frac{1}{8} \end{pmatrix} = \begin{pmatrix} \frac{3a}{16} - \frac{1}{8} & \frac{a}{8} + 2b \\ 0 & -3b - \frac{1}{8} \end{pmatrix}.\]We want this to be the identity matrix, so $ \frac{3a}{16} - \frac{1}{8} = 1, $ $ \frac{a}{8} + 2b = 0, $ and $ -3b - \frac{1}{8} = 1 $. Solving, we find $ (a,b) = \boxed{\left( 6, -\frac{3}{8} \right)} $.