yes, there is the Sherman-Morrison formula
$$\det B=(\det A)(b-yA^{-1}x),$$
where $b, x$ and $y$ are blocks:
$$B=\begin{pmatrix} A & x \\ y & b \end{pmatrix}.$$

*Edit*. After Hachino's comment, one can also write
$$\det B=b\det A-y\hat Ax,$$
where $\hat A$ is the transpose of the cofactor matrix.