Let $M$ be a symmetric real matrix $(n+1)$ by $(n+1)$. Let the $n$ by $n$ top left corner matrix be $A$ and denote $c$ to be the column vector with $n$ entries and the $(n+1, n+1)$ of $M$ to be $x$, so that $$ M = \begin{bmatrix} A && c \\ c^T && x \end{bmatrix}. $$ We know that the $\det A \neq 0$.
I want to show (I think this is true!) that $$ x - c^T A^{-1} c = \frac{\det M}{\det A}. $$
For $n=1$, it is easy enough to check the statement is true. If someone could have any idea on how I can prove this, it would be appreciated! Thank you!
My neighbour showed me a nice answer so I decided to post it here. First notice that $$ \begin{bmatrix} A^{-1} && 0 \\ 0 && 1 \end{bmatrix} M = \begin{bmatrix} I && A^{-1} c \\ c^T && x \end{bmatrix}. $$ If we row deduce the last matrix we can get $$ \begin{bmatrix} I && A^{-1} c \\ 0 && x - c^T A^{-1} c \end{bmatrix}. $$ Thus taking the determinant of both sides, taking into account the row reduction we get $$ \det M/\det A = x - c^T A^{-1} c. $$