I'm trying to prove the following: Let $A$ be a $k\times k$ matrix, let $D$ have size $n\times n$, and $C$ have size $n\times k$. Then,
$$\det\left(\begin{array}{cc}
A&0\\
C&D
\end{array}\right) = \det(A)\det(D).$$
Can I just say that $AD – 0C = AD$, and I'm done?
Best Answer
If $A$ is singular, its rows are linearly dependent, hence the rows of the entire matrix are linearly dependent, hence boths sides of the equation vanish.
If $A$ is not singular, we have
$$\pmatrix{I&0\\-CA^{-1}&I}\pmatrix{A&0\\C&D}=\pmatrix{A&0\\0&D}\;.$$
The determinants of the two new matrices are perhaps easier to derive from the Laplace expansion than that of the entire matrix. They are $1$ and $\det A \det D$, respectively, and the result follows.
Another way to express this is that if $A$ is not singular you can get rid of $C$ by Gaussian elimination.