<s>[INST] Let $M=\big( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \big)$. Prove $\det(M)=\det(A)\cdot \det(D-C·A^{-1}·B)$. Let  (shown in matrix blocks) $M=\big( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \big)$ a square matrix such that $A$ is invertible and $D$ is a square matrix. I have to prove that
$$\det(M)=\det(A)\cdot \det(D-C·A^{-1}·B)$$
I also have an indication to use it:
Consider previously these cases:
\begin{align*}
A_1 &= \begin{pmatrix} A & 0 \\ 0 & I \end{pmatrix},\\
A_2 &= \begin{pmatrix} I & 0 \\ 0 & D \end{pmatrix},\\
A_3 &= A_1·A_2,\\
A_4 &= \begin{pmatrix} I & A^{-1}·B \\ 0 & I \end{pmatrix},\text{ and}\\
A_5 &= A_3\cdot A_4.
\end{align*} 
The problem is difficult in general for me and I don't even know how to use the indication. 
 [/INST] The indication shows that 
$\det\begin{pmatrix}A&B\\0&D\end{pmatrix}=\det(A)\det(D)$.
Similarly $\det\begin{pmatrix}A&0\\C&D\end{pmatrix}=\det(A)\det(D)$.
Now $\begin{pmatrix}I&0\\-CA^{-1}&I\end{pmatrix}\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&B\\0&D-CA^{-1}B\end{pmatrix}$
It follows from the indication that $\det(I)\det(I)\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(A)\det(D-CA^{-1}B)$
</s>