# If $\operatorname{rank}\left( \begin{bmatrix} A &B \\ C &D \end{bmatrix}\right)=n$ prove that $\det(AD)=\det(BC)$

Let $A,\ B,\ C,\ D \in \mathcal{M}_n(\mathbb{C})$. If $\operatorname{rank}\left( \begin{bmatrix} A &B \\ C &D \end{bmatrix}\right)=n$, prove that $\det(AD)=\det(BC)$.

#### Solutions Collecting From Web of "If $\operatorname{rank}\left( \begin{bmatrix} A &B \\ C &D \end{bmatrix}\right)=n$ prove that $\det(AD)=\det(BC)$"

Assume first that the first $n$ columns are independent. Then the last $n$ columns are linear combinations of the first $n$. Therefore we have the equality

$$\begin{bmatrix} A \\ C \end{bmatrix} \cdot V = \begin{bmatrix} B \\ D \end{bmatrix}$$

so we’re done.

Now, if the first $n$ columns are not linearly independent, both sides are $0$.

Obs: to understand what $V$ is: the first column of $V$ consists of the coefficients in the writing of the first column of $\begin{bmatrix} B \\ D \end{bmatrix}$ as a linear combination of the columns of $\begin{bmatrix} A \\ C \end{bmatrix}$.