Construct a matrix $M$ from $A$ and $B$ such that $\det(M)=\det(A)-\det(B)$

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Given two $n \times n$ symmetric matrices $A$ and $B$, is there a generic way to construct a larger block matrix $M$ such that $\det(M) = \det(A) - \det(B)$?

A simple block expression is desired, in the sense that the block components of $M$ are constant matrices or obtained by solving matrix equations involving $A$ and $B$. Constructions of $M$ involving short algebraic expressions for its components in terms of the components of $A$ and $B$ would also be interesting, but not something that expands to an exponential number of terms in the components of $A$ and $B$ like just sticking $\det(A)$ in as a term of $M$.

If this is not possible in the general case, what restrictions can be placed on $A$ and $B$ to make this possible?

The only case I know of is when the difference of $B$ and $A$ can be written as a product of a column vector $C$ and its transpose: $B-A = CC^T$. This allows us to construct a matrix $M$ such that:

$$ M = \begin{bmatrix} A & C \\ C^T & 0 \end{bmatrix} $$ $$ \det(M) = \det(A) - \det(A + CC^T) = \det(A) - \det(B) $$

I'm curious if there is some way to construct an appropriate block matrix to make this possible for arbitrary symmetric matrices $A$ and $B$.

The first comment to this question
Find a matrix with determinant equals to $\det{(A)}\det{(D)}-\det{(B)}\det{(C)}$
suggests the answer is trivial by choosing $$M = \begin{bmatrix} A & B \\ I & I \end{bmatrix}$$ but that doesn't appear to work when I tried some numerical examples.