Complementary minor of a kernel of given full-rank matrix $A$ is equal to minor of $A$.

Question

Let $d \leq n$, $d,n \in\mathbb{N}$. Suppose $\begin{pmatrix}A & B \end{pmatrix},\begin{pmatrix}C \\ D \end{pmatrix}$ be a block matrix with $A \in Mat_{\mathbb{Z}^{d\times d}},B \in Mat_{\mathbb{Z}^{d\times (n-d)}}, C \in Mat_{\mathbb{Z}^{d\times (n-d)}}, D \in Mat_{\mathbb{Z}^{(n-d)\times (n-)d}}$ such that 1) $\begin{pmatrix}A & B \end{pmatrix}$ has rank $d$ and 2) $$\begin{pmatrix}A & B \end{pmatrix}\begin{pmatrix}C \\ D \end{pmatrix} = 0 $$ which is equivalent to say $$AC + BD=0.$$ Also suppose that each columns of $\begin{pmatrix}C \\ D \end{pmatrix}$ consists of a $\mathbb{Z}$-basis of $\ker \begin{pmatrix}A & B \end{pmatrix}$. Then can we say $\det(A) = \det(D)$? It is mentioned in some book as a basic fact of linear algebra, but I don't know how to prove it. Any hint will be appreciated.

Answer 1

Take $d=1$ and $n=2$; let $A=3$, $B=0$, $C=0$, $D=1$. These satisfy the hypotheses, yet $\det A\not= \det D$.