# How to prove the simplified form of $det\left( \begin{matrix} A& \vec{x}\\ \vec{y}^T& a\\ \end{matrix} \right)$?

The question is, how to prove $$det\left( \begin{matrix} A& \vec{x}\\ \vec{y}^T& a\\ \end{matrix} \right) =a\cdot det\left( A \right) -\vec{y}^T\left( adj\left( A \right) \right) \vec{x}$$, where $$A$$ is a square matrix, $$\vec y,\vec x$$ are vectors and $$a\in\mathbb R$$?

There's where I go for now: When $$A$$ is nonsingular, it's very easy to see the result with the help of schur complement $$S$$, i.e., $$det\left( \begin{matrix} A& \vec{x}\\ \vec{y}^T& a\\ \end{matrix} \right) =det\left( A \right) det\left( S \right) \\ =det\left( A \right) det\left( a-\vec{y}^TA^{-1}\vec{x} \right) \\ =det\left( A \right) \left( a-\vec{y}^TA^{-1}\vec{x} \right) \\ =a\cdot det\left( A \right) -\vec{y}^T\left( adj\left( A \right) \right) \vec{x}$$ But the problem is, when $$A$$ is singular, I don't know how to get the result.

Let $$A_i$$ be the matrix obtained by deleting row $$i$$ from $$A$$ and putting $$y$$ in the bottom row. Then from the Laplace expansion of the determinant along the last column, we have that the determinant of the full matrix is $$(-1)^{n+1}\sum_{i=1}^n (-1)^i x_i\det A_i + a\cdot\det A$$
So what is $$\det(A_i)$$? Letting $$M_{ij}(A)$$ be the minors of $$A$$ and Laplace expanding along the bottom row gives $$(-1)^{n}\sum_{j=1}^n(-1)^jy_jM_{ij}(A) = (-1)^{n+i}\sum_{j=1}^n y_j[\operatorname{adj}A]_{ji}$$ Putting these together gives $$(-1)^{n+1}\sum_{i = 1}^n(-1)^i x_i (-1)^{i+1}\sum_{j=1}^n y_j[\operatorname{adj}A]_{ji} = -\sum_{i = 1}^n\sum_{j=1}^n y_j [\operatorname{adj}A]_{ji}x_i = -y^T (\operatorname{adj}A) x$$ So the whole determinant is $$a\cdot\det A -y^T (\operatorname{adj}A) x$$.
Find a sequence $$\{A_n\}_n$$ of nonsingular matrices converging to $$A$$, apply your result to each $$\left( \begin{matrix} A_n& \vec{x}\\ \vec{y}^T& a\\ \end{matrix} \right),$$ and then take the limit as $$n\to\infty$$ on both sides.