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.