$\det{\begin{bmatrix} A & B \\ C & D\end{bmatrix}}=(D-1)\det(A) +\det(A-BC) $?

Question

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!

Answer 1

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 $1 - \mathrm r^T \mathrm A^{-1} \mathrm c = \det(\mathrm I_n - \mathrm A^{-1} \mathrm c \mathrm r^T)$.

Answer 2

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 $D = (D + 1) - 1$.