$\det{\begin{bmatrix} A & B \\ C & D\end{bmatrix}}=(D-1)\det(A) +\det(A-BC) $? I found this identity in Wikipedia,

When $D$ is a $1 \times 1$ matrix, $B$ is a column vector, and $C$ is a row vector, then
$$\det{\begin{bmatrix} A &  B \\  C & D\end{bmatrix}}=(D-1)\det(A) +\det(A-BC)= (D+1)\det(A) -\det(A+BC)$$

I wonder how to prove that. Thanks!
 A: Observe that
\begin{equation*}
\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
= \begin{bmatrix}
A & B + 0\\
C & (D-1) + 1
\end{bmatrix}.
\end{equation*}
Therefore, using multilinearity of determinants
\begin{equation*}
\det\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
= \det \begin{bmatrix}
A & 0\\
C & D - 1
\end{bmatrix} + 
\det \begin{bmatrix}
A & B\\
C & 1
\end{bmatrix}.
\end{equation*}
Now, the first determinant on the RHS is equal to $\det(A) \det(D  -1) = (D - 1) \det(A)$, and the second determinant is $\det([1])\det(A - B[1]^{-1}C) = \det(A - BC)$, from which the first form of the result follows. The second form is obtained similarly by writing $D = (D + 1) - 1$.
A: Let
$$\mathrm M := \begin{bmatrix} \mathrm A & \mathrm c\\ \mathrm r^T & \alpha\end{bmatrix}$$
where $\mathrm A \in \mathbb R^{n \times n}$ and $\mathrm c, \mathrm r \in \mathbb R^n$. Assuming that $\mathrm A$ is invertible,
$$\underbrace{\begin{bmatrix} \mathrm I_n & 0_n\\ -\mathrm r^T \mathrm A^{-1} & 1\end{bmatrix}}_{\det (\cdot) = 1} \begin{bmatrix} \mathrm A & \mathrm c\\ \mathrm r^T & \alpha\end{bmatrix} = \begin{bmatrix} \mathrm A & \mathrm c\\ 0_n^T & \alpha - \mathrm r^T \mathrm A^{-1} \mathrm c\end{bmatrix}$$
Hence,
$$\begin{array}{rl} \det (\mathrm M) &= \det \begin{bmatrix} \mathrm A & \mathrm c\\ 0_n^T & \alpha - \mathrm r^T \mathrm A^{-1} \mathrm c\end{bmatrix}\\\\ &= \det (\mathrm A) \cdot (\alpha - \mathrm r^T \mathrm A^{-1} \mathrm c)\\\\ &= (\alpha - 1) \cdot \det (\mathrm A) + \det (\mathrm A) \cdot (1 - \mathrm r^T \mathrm A^{-1} \mathrm c)\\\\ &= (\alpha - 1) \cdot \det (\mathrm A) + \det (\mathrm A) \cdot \det(\mathrm I_n - \mathrm A^{-1} \mathrm c \mathrm r^T)\\\\ &= (\alpha - 1) \cdot \det (\mathrm A) + \det(\mathrm A - \mathrm c \mathrm r^T)\end{array}$$
where we used Weinstein-Aronszajn to conclude that $1 - \mathrm r^T \mathrm A^{-1} \mathrm c = \det(\mathrm I_n - \mathrm A^{-1} \mathrm c \mathrm r^T)$.
