I came across a question on SE Math and was reading a proof on Sylvester's Determinant Theorem posted by anon. But I have a few doubts as I'm reading it. I will re-produce from the pdf file on the part that I don't understand so that it makes it easier for reference.
In Theorem 9, it says if matrix $A$ and $D$ are invertible, then in the block matrix below, $ \begin{vmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{vmatrix} = \left | A \right |\left | D-CA^{-1}B \right | = \left | D \right |\left | A-BD^{-1}C \right | $
The proof given to the claim was: $\begin{align*} \begin{bmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{bmatrix} &= \begin{bmatrix} A_{m \times m} & 0_{m \times n}\\ C_{n \times m} & I_{n \times n} \end{bmatrix} \begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ C_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix}\\ &= \begin{bmatrix} I_{m \times m} & B_{m \times n}\\ 0_{n \times m} & D_{n \times n} \end{bmatrix} \begin{bmatrix} A-BD^{-1}C_{m \times m} & 0_{m \times n}\\ D^{-1}C_{n \times m} & I_{n \times n} \end{bmatrix} \end{align*}$
First, I don't understand how $\begin{bmatrix} A_{m \times m} & 0_{m \times n}\\ C_{n \times m} & I_{n \times n} \end{bmatrix} \begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ C_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix}$ was derived from $\begin{bmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{bmatrix}$.
Second, when I did a multiplication between the matrices, $ \begin{align*} &\begin{bmatrix} A_{m \times m} & 0_{m \times n}\\ C_{n \times m} & I_{n \times n} \end{bmatrix} \begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ C_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix} =\\ &\begin{bmatrix} A_{m \times m}I_{m \times m}+0_{m \times n}C_{n \times m} & A_{m \times m}A^{-1}B_{m \times n}+0_{m \times n}(D-CA^{-1}B_{m \times n})\\ C_{n \times m}I_{m \times m}+I_{m \times m}C_{n \times m} & C_{n \times m}A^{-1}B_{m \times n}+I_{n \times n}(D-CA^{-1}B_{m \times n}) \end{bmatrix}=\\ &\begin{bmatrix} A_{m \times m} & B_{m \times n}\\\ 2C_{n \times m} & D_{n \times n} \end{bmatrix}\neq \begin{bmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{bmatrix} \end{align*}$ It's weird that in my multiplication doesn't get back the original matrix.
How was the first equation being derived by sort of "splitting" the matrix into 2 matrices? Also, why doesn't my multiplication get back the original matrix?
Thanks for any help.