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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.

Any idea? Thanks in advance!

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    $\begingroup$ You could simply use limits and the continuity of the determinant $\endgroup$ Commented Jun 6, 2022 at 13:07

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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 $a\cdot\det A -y^T (\operatorname{adj}A) x$.

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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.

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