<s>[INST] $\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!
 [/INST] 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 $D = (D + 1) - 1$.
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