Determinant of block matrix with A size 1x1

Question

I have the following block matrix with dimensions

$ \begin{bmatrix} A & B\\ C & D \end{bmatrix} == \begin{bmatrix} (1\times1) & (1\times n)\\ (n\times1) & (n\times n) \end{bmatrix}$

I'm trying to compute the determinant with the following formula

$Det[A]Det[D-CA^{-1}B]$

which is given here on Wikipedia.

So, for my question. Since "A" is given as a 1x1 in my problem, would that mean that my equation for the determinant would reduce to

$Det[AD-CB]$

I will note that the inverse of "A" does exist. If it didn't, obviously this entire approach is wrong. I'm just not sure I can distribute $Det[A]$ into $Det[D-CA^{-1}B]$ as freely as I'd like to?

Answer 1

No. $A = a$ is a number. So you have for your block matrix $X$ (if you applied the Wiki formula correctly): $$ Det[X] = Det[A]Det[D-CA^{-1}B] = aDet[D-Ca^{-1}B] = aDet[a^{-1}(AD-CB)] = aa^{-n}Det[AD-CB] = a^{1-n}Det[AD-CB]. $$