# Determinant of block matrix with A size 1x1

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?

• The answer is Yes! – mike Feb 24 '16 at 6:40
• @mike Friedrich says "No"? – ThatsRightJack Feb 24 '16 at 6:46
• @mike: Try some simple examples. Here is one: $\left\begin{matrix}2 & 1 & 1\\1 & 1 & 1\\1 & 0 & 1\end{matrix}\right)$. I don't see why the system doesn't take this one. – Friedrich Philipp Feb 24 '16 at 7:04
• Sorry. I did not pay attention to the scalars. – mike Feb 25 '16 at 18:06

No. $A = a$ is a number. So you have for your block matrix $X$ (if you applied the Wiki formula correctly): $$Det[X] = Det[A]Det[D-CA^{-1}B] = aDet[D-Ca^{-1}B] = aDet[a^{-1}(AD-CB)] = aa^{-n}Det[AD-CB] = a^{1-n}Det[AD-CB].$$
• Wow...good thing I asked. Just to clarify. What is $a^{-n}$ – ThatsRightJack Feb 24 '16 at 6:44
• $a^{-n} = 1/a^n$. I recall that $a$ is a number. BTW: The block in the upper right is 1xn and not nx1. – Friedrich Philipp Feb 24 '16 at 6:46
• Thanks for pointing that out. I fixed it in the question. I'm still a bit confused by your answer with $a^{-n}$. Wouldn't "n" just be one? Hence $aa^{-1}=1$ and the it does reduce to $Det[AD-CB]$? – ThatsRightJack Feb 24 '16 at 6:56
• No. It is YOUR $n$. Namely the size of the matrix $D$. Note that $Det$ is not linear. For example, $Det(2I) = Det\left(\begin{matrix}2 & 0\\0 & 2\end{matrix}\right) = 4 = 2^2Det(I)\neq 2Det(I)$, where $I$ is the 2x2 identity matrix. In general, you have $Det(tA) = t^nDet(A)$, where $n$ is the size of $A$. – Friedrich Philipp Feb 24 '16 at 6:59