Prove $\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A-B)\det(A+B)$, even when $A$ and $B$ are not commutative. I am aware of the following identity
$\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(A)\det(D - CA^{-1}B)$
When $A = D$ and $B = C$ and when $AB = BA$ the above identity becomes
$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A)\det(A - BA^{-1}B) = \det(A^2 - B^2) = \det(A-B)\det(A+B)$.
However, I couldn't prove this identity for the case where $AB \neq BA$.
EDIT: Based on @Trebor 's suggestion.
I think I could do the following.
$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = 
\det\begin{bmatrix}A & B\\ B-A & A-B\end{bmatrix} = \det(A^2-B^2) = \det(A-B)\det(A+B)$.

$\begin{pmatrix} A & B \\ B & A \end{pmatrix}\xrightarrow{\text{row1 -= row2}}\begin{pmatrix} A-B & B-A \\ B & A\end{pmatrix}\xrightarrow{\text{col2 += col1}}\begin{pmatrix} A-B & O  \\ B & A+B\end{pmatrix}$
