Calculating the determinant of a matrix using its rank

Question

Let A, B, C and D be real n×n matrices. If

$$\operatorname{rank} \begin{bmatrix} \ A & B \\[0.3em] \ C & D \\[0.3em] \end{bmatrix} = n$$

then show that

$$\det \begin{bmatrix} \det A & \det B \\[0.3em] \det C & \det D \\[0.3em] \end{bmatrix} = 0$$

Answer 1

Consider two cases.

1. $\def\rk{\operatorname{rank}}\rk([A~B])<n$. Then $\det(A)=\det(B)=0$ and the result is obvious.

2. $\rk([A~B])=n$. Now the fact that adding the rows of $[C~D]$ to the matrix does not increase the rank means the every such row is a linear combination of the rows of $[A~B]$. This means that there is some $n\times n$ matrix $X$ such that $[C~D]=X[A~B]$, which means $C=XA$ and $D=XB$. Now using $\det(C)=\det(X)\det(A)$ and $\det(D)=\det(X)\det(B)$ the result again follows easily.