For $A, B, C, D \in \mathbb{F}^{n \times n}$, show that $\det{\begin{pmatrix} A & B \\ C & D\end{pmatrix}} = \det{A}\det{D} - \det{B}\det{C}$

Question

I found this exercise in the linear algebra script of my university.

For $A, B, C, D \in \mathbb{F}^{n \times n}$, show that

$$\det{\begin{pmatrix} A & B \\ C & D\end{pmatrix}} = \det{A}\det{D} - \det{B}\det{C}$$

While I know this would be true for $C = 0$, in that case we can bring $A$ and $D$ in row echelon form. But how about the case where $C \neq 0$?

As an example I tried the matrix $\begin{pmatrix}2 & 3 & 1 & 0 \\ 3 & 2 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{pmatrix}$, which should have a determinant of $-6$ if the lemma was true. But the determinant of the matrix is $-8$.

Can I conclude that the exercise has a typo?

Answer 1

Here is a route you could take.

You have rightly asserted that if we have $$\det\begin{pmatrix}A&B\\0&D\end{pmatrix}=\det(A)\det(D)$$.

Similarly one can trivially write $$\det\begin{pmatrix}A&0\\C&D\end{pmatrix}=\det(A)\det(D)$$

Now observe that $$\begin{pmatrix}I&0\\-CA^{-1}&I\end{pmatrix}\begin{pmatrix}A&B\\C&D\end{pmatrix}=\begin{pmatrix}A&B\\0&D-CA^{-1}B\end{pmatrix}$$

Thus one can write that $$\det(I)\det(I)\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(A)\det(D-CA^{-1}B)$$

Answer 2

The correct result is $$\det\begin{bmatrix} A & B \\ C & D\end{bmatrix}= \det(AD-BC)$$

https://en.wikipedia.org/wiki/Block_matrix