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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?

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    $\begingroup$ Try diagonal matrices or just $1\times 1$-matrices: The formula seems to be just wrong. $\endgroup$ – Peter Melech Mar 6 '18 at 10:40
  • $\begingroup$ That seems to be "a formula" that could, perhaps, work for very, very particular cases... $\endgroup$ – DonAntonio Mar 6 '18 at 10:43
  • $\begingroup$ I figured as such that's why I wanted to hunt it down. Are there any identities that do a similar expansion? i.e. that expand $\det(A+B)$ into $\det(A)+\det(B)+f(A,B)$ where $f$ is the "correction" from non-distributivity? $\endgroup$ – Jepsilon Mar 6 '18 at 10:49
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    $\begingroup$ The formula is true for $2\times2$ matrices, but false for larger ones (or smaller ones, as Peter Melech has pointed out in his comment above). $\endgroup$ – user1551 Mar 6 '18 at 11:24
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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

$$\begin{array}{rl} \det (\mathrm A + \mathrm B) &= \det \left( \mathrm A \left( \mathrm I_2 + \mathrm A^{-1} \mathrm B \right) \right)\\ &= \det (\mathrm A) \cdot \det \left(\mathrm I_2 + \mathrm A^{-1} \mathrm B \right)\\ &= \det (\mathrm A) \cdot \left( 1 + \det(\mathrm A^{-1} \mathrm B) + \mbox{tr}(\mathrm A^{-1} \mathrm B) \right)\\ &= \det (\mathrm A) + \det (\mathrm A) \cdot \det(\mathrm A^{-1} \mathrm B) + \mbox{tr} \left( \det (\mathrm A) \, \mathrm A^{-1} \mathrm B \right)\\ &= \det (\mathrm A) + \det (\mathrm B) + \mbox{tr} \left( \mbox{adj}(\mathrm A) \mathrm B \right)\end{array}$$

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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 $(A^{(s)})^{-1}$

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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.

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