Formula for determinant of sum of matrices Some time ago I came across this apparently quite obscure formula that expands the determinant of a sum of two matrices that I had put on my notes (assuming that I made no errors in my writing):
$$\det(A+B)=\det(A)+\det(B)+\text{Tr}(\text{adj}(A)B)$$
Where $\text{adj}()$ denotes the adjugate of the matrix. I cannot seem to find any mention of this formula online. Does anyone know of the name (and maybe a proof) of it? Furthermore, is there any more info on it, like conditions that $A$ and $B$ must obey for it to hold?
 A: It seems to me that you are asking for is a sort of "Taylor expansion" keeping track of the "errors" of various weights. So perhaps the generalisation you want is this, which is not hard to check:
$$
\det(A+xB)=
\det A \sum_{s} x^s\ \textrm{tr}\left( (A^{-1})^{(s)} B^{(s)}\right)
$$
where $X^{(s)}$ is the matrix of $s\times s$ 
cofactors. 
There are variants got by expressing $A^{-1}$ in terms of the determinant and adjugate, or replacing $(A^{-1})^{(s)}$ by $(A^{(s)})^{-1}$
A: For $2 \times 2$ matrices, the following holds
$$\det (\mathrm I_2 +  \mathrm M) = 1 + \det(\mathrm M) + \mbox{tr}(\mathrm M)$$
If $\rm A, B$ are $2 \times 2$ matrices, and temporarily assuming that $\rm  A$ is invertible, then
$$\begin{array}{rl} \det (\mathrm A +  \mathrm B) &= \det \left( \mathrm A \left( \mathrm I_2 +  \mathrm A^{-1} \mathrm B \right) \right)\\ &= \det (\mathrm A) \cdot \det \left(\mathrm I_2 +  \mathrm A^{-1} \mathrm B \right)\\ &= \det (\mathrm A) \cdot \left( 1 + \det(\mathrm A^{-1} \mathrm B) + \mbox{tr}(\mathrm A^{-1} \mathrm B) \right)\\ &= \det (\mathrm A) + \det (\mathrm A) \cdot \det(\mathrm A^{-1} \mathrm B) +  \mbox{tr} \left( \det (\mathrm A) \, \mathrm A^{-1} \mathrm B \right)\\ &= \det (\mathrm A) +  \det (\mathrm B) + \mbox{tr} \left( \mbox{adj}(\mathrm A) \mathrm B \right)\end{array}$$
A: Given two $2 \times 2$ matrices $A$ and $B$,
$$\det(A + B) = 2(\det(A) + \det(B)) - \det(A - B)$$
This can be proven directly via the formula for the determinant of a $2 \times 2$ matrix, and therefore applies whether the matrices are invertible or not.
If $A = \begin{bmatrix}a&c\\b&d\end{bmatrix}$ and 
$B = \begin{bmatrix}e&g\\f&h\end{bmatrix}$,
$$\text{tr}(\text{adj}(A)B) = \text{tr}\left(\begin{bmatrix}d&-c\\-b&a\end{bmatrix} \begin{bmatrix}e&g\\f&h\end{bmatrix}\right) = \text{tr}\left(\begin{bmatrix}de-cf&dg-ch\\af-be&ah-bg\end{bmatrix}\right) = de-cf+ah-bg$$
$$\det(A) + \det(B) - \det(A-B) = ad - bc + eh - fg - (a-e)(d-h) + (b-f)(c-g)\\$$
Combining like terms shows that $$\text{tr}(\text{adj}(A)B) = \det(A) + \det(B) - \det(A - B)$$
and therefore that $$\det(A) + \det(B) + \text{tr}(\text{adj}(A)B) = 2(\det(A) + \det(B)) - \det(A - B) = \det(A + B)$$
To answer your other questions, I don't know the name of either formula (I discovered the one I used independently, though I doubt I'm the first). And like I said, my formula applies to all $2\times2$ matrices, invertible or not, and your formula is equivalent to mine, so the matrices need not be of a special form.
