I am attempting to complete the exercises in my textbook on matrix differential calculus. This question is giving me some trouble. These are the problems I am attempting to complete:
If $|A| \ne 0$, prove that
$$\begin{vmatrix} A & b \\ a' & \alpha\end{vmatrix}=(\alpha - a'A^{-1}b)|A|.$$
If $\alpha \ne 0$, prove that
$$\begin{vmatrix} A & b \\ a' & \alpha\end{vmatrix}=\alpha |A-\left( \frac1\alpha\right)ba'|$$
I found on wikipedia Determinant of Block Matrix which shows how if you have a partitioned matrix you can decompose that matrix into an upper and lower triangular matrix and apply the product rule to the determinant to find it. Seen as follows:
When $A$ is invertible, one has $$\det\begin{pmatrix} A & B \\ C & D \end{pmatrix}=\det(A) \times \det(D-CA^{-1}B)$$ as can be seen by employing the decomposition
$$\begin{pmatrix} A & B \\ C & D\end{pmatrix}=\begin{pmatrix} A & 0 \\ C & I_m\end{pmatrix}\begin{pmatrix} I_n & A^{-1}B \\ 0 & D-CA^{-1}B\end{pmatrix}$$
My question is how to extend this logic to a partitioned matrix where on wikipedia's example, $B$ and $C$ are vectors. Down the page a bit they state
When $D$ is a $1 \times 1$ matrix, $B$ is a column vector and $C$ is a row vector then
$$\det\begin{pmatrix} A & B \\ C & D \end{pmatrix}=\det(D-CA^{-1}B) \det(A)$$
which is essentially what I am trying to prove. I am just not quite sure how to prove this. I have attempted to use laplace expansion, but that didn't really get me anywhere. Not sure if this is because I am missing something, or if laplace expansion is not the proper technique to prove this.
Row and column matrices are special case of general matrices, the same rules apply.
Suppose you have accepted the fact that
$$\det\begin{pmatrix} A & B \\ C & D\end{pmatrix}=\det(A) \det(D-CA^{-1}B)$$
Then by letting $B=b$, $C=a'$ and $D=\alpha$,
We have if $\det(A) \ne 0$,
$$\det\begin{pmatrix} A & b \\ a' & \alpha\end{pmatrix}=\det(A) \det(\alpha-a'A^{-1}b)=(\alpha - a'A^{-1}b) |A|$$
If $\alpha \ne 0$,
\begin{align}\det\begin{pmatrix} A & b \\ a' & \alpha\end{pmatrix}&=\det \left( \begin{pmatrix} 0^T & 1 \\ I_{n-1} & 0\end{pmatrix}\begin{pmatrix} \alpha & a' \\ b & A\end{pmatrix} \begin{pmatrix} 0^T & 1 \\ I_{n-1} & 0\end{pmatrix}^T\right)\\&=\det \left( \begin{pmatrix} 0^T & 1 \\ I_{n-1} & 0\end{pmatrix}\right)^2 \det\left(\begin{pmatrix} \alpha & a' \\ b & A\end{pmatrix} \right)\\&=\det\left(\begin{pmatrix} \alpha & a' \\ b & A\end{pmatrix} \right)\\ &=\alpha \det(A-b\left(\frac1\alpha\right) a')\end{align}