Block matrix determinant with symmetrically placed blocks

Question

I am trying to solve the determinant of the following block matrix

$$\begin{bmatrix}A-Ia&B\\B &A-Ib \end{bmatrix}$$

where $a$ and $b$ are integers and $I$ is an identity matrix. Matrices $A$ and $B$ are square.

Unfortunately, I can't invoke the general expression since I don't want to attempt to invert any of these matrices (very complex). I see from Wikipedia that

$$\det\begin{bmatrix}A&B\\B &A \end{bmatrix} = \det(A+B)\det(A-B)$$ Can anyone help me prove that this can be applied to my matrix (Such that I might get something like $\det(A-Ia+B)\det(A-Ib-B)$)? Or alternatively, can anyone solve the determinant of my matrix without taking an inverse?

Thanks!

Answer 1

The proof for the Wikipedia formula was given in this thread, which unfortunately still leads to a system for which an inverse needs to be taken in order to solve the determinant. At least I was able to prove that in this general case it's impossible:

Inserting my matrix into the proof for \begin{bmatrix}A&B\\B &A \end{bmatrix}

Gives

$$\begin{bmatrix}I&-I\\0 &I \end{bmatrix} \begin{bmatrix}A-Ia&B\\B &A-Ib \end{bmatrix} \begin{bmatrix}I&I\\0 &I \end{bmatrix} = \begin{bmatrix}A-Ib-B&Ib-Ia\\B &A-Ib+B \end{bmatrix}$$

Which still requires an inverse to solve, given that none of the appropriate sub-matrices commute.

Answer 2

Hint : If $(A-aI)$ be invertible then $$\det\begin{bmatrix}A-Ia&B\\B &A-Ib \end{bmatrix}=\det(A-aI)\det(A-bI-B(A-aI)^{-1}B)$$