Let $A,B,C$ and $D$ be $n \times n$ matrices such that $AC = CA$. Prove that $$\det \begin{pmatrix} A & B\\ C & D\end{pmatrix} = \det(AD-CB)$$
The solution is to first assume that $A$ is invertible and then consider the product
$$\begin{pmatrix}
I & O\\
-CA^{-1} & I
\end{pmatrix}\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}=\begin{pmatrix}
A & B\\
O & D-CA^{-1}B
\end{pmatrix}$$
then it is not hard to prove that the claim is true if $A$ in invertible. Finally, we use the fact that the set $GL_n$ form a dense open subset of $M_n$ to get rid of the invertibility assumption. My question is: how to come up with such a weird matrix $\begin{pmatrix}
I & O\\
-CA^{-1} & I
\end{pmatrix}$? thank you so much. Is there any other problems that uses the technique of assuming invertibility? (one of which i know is to prove $\det (I+AB) = \det(I+BA)$), thanks in advance
Best Answer
As the others have pointed out, the block matrix originates form the elementary matrix in Gaussian elimination. The ultimate goal here is to transform the matrix in question into a block-triangular form. Yet, in general, such uses of block matrices serve different goals. So, rather than explaining how to devise the block matrix in this specific case, I think it is more important to note that you can sometimes simplify your problem by using a transformation matrix in block form.
Your proof actually has three features that are quite common in various proofs:
I think it is worthwhile to put all these three tricks in your toolbox: