Calculate determinant of $ M=\left( \begin{array}{cc} A&-\vec d^T \\ \vec c& b \\ \end{array} \right) $. I have a block matrix
$$
M=\left(
  \begin{array}{cc}
     A&-\vec d^T  \\
     \vec c& b \\
  \end{array}
\right)
$$
where $A$ is a $(n-1)\times (n-1)$ matrix, $\vec d,\ \vec c$ are two vectors of dimension $=n-1$, while $b$ is a real number. Thus matrix $M$ is a $n\times n$ square matrix. My question is as follows. Knowing $\det A$, $\vec d,\ \vec c$ and $b$, is it possible to calculate $\det M$?
Thanks in advance.
 A: Here $c,d$ are rows and not vectors. There is a formula that is valid for any $b$ (the last equality in kubek's proof is false). According to [Horn, Johnson "Matrix Analysis" (2013), 0.8.5.10 ], one has $\det(M)=b\det(A)+c. adj(A).d^T$, where $adj(A)$ is the classical adjoint of $A$.
A: There is a general fact, that when we have:
$$M=\begin{bmatrix} A & B\\ C & D \end{bmatrix},$$
where $A\in \mathbb{R}^{p\times p}, D \in \mathbb{R}^{q\times q}, B \in \mathbb{R}^{p\times q},C\in \mathbb{R}^{q\times p}$ and $A$ is invertible, then we have:
$$M=\begin{bmatrix} I & O \\ CA^{-1} & I \end{bmatrix} \cdot \begin{bmatrix} A & B\\ O & D-CA^{-1} B \end{bmatrix}.$$
From this and the formula for determinant of block upper and lower diagonal matrices we get:
$$\det M=\det A \cdot \det (D-CA^{-1}B).$$
In the similar way we can get the following formula, when $D$ is invertible:
$$\det M=\det D \cdot \det (A-BD^{-1}C).$$
So here in your case, when $b\neq 0,$ then you have:
$$\det M=b \det(A+d^Tb^{-1}c)=b^{1-p}\det(bA+d^Tc).$$
