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Suppose we have some arbitrary square matrices $ A,B $ and $C$ of dimension n. We can find the determinant of their sum, i.e. $$ y = \det(A+B+C) $$ My question is, how does the matrix $A$ effect the value of $y$? In other words, is it possible to know how much the value of $y$ will change w.r.t. to addition or removal of $A$.

My approach: I added $\phi_i$ (it is either 0 or 1) to the equation and tried to differentiate it w.r.t. $\phi_i$.

$$ y = \det(\phi_A A + \phi_B B + \phi_C C) $$ $$ \dfrac{\partial y}{\partial \phi_A} = \text{tr}(\text{adj}(y)A) $$ But I am unable to proceed any further from this. I don't know whether this result is correct or am I going in the right direction. What should I do?

Note: Take n = 2 or 3. Keep it small.

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  • $\begingroup$ $y$ is a number, not a matrix, so $\text{adj}(y)$ makes no sense. $\endgroup$
    – Robert Israel
    Dec 11, 2019 at 14:16

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First of all, we may as well combine $B$ and $C$ in a single matrix, so we consider $ y = \det(tA+B)$. Then $$ \dfrac{d}{dt} \det(tA+B) = \text{tr}(\text{adj}(tA+B)A) $$ When $tA+B$ is invertible, this is $$ \det(tA+B)\; \text{tr}\left((tA+B)^{-1} A\right)$$

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  • $\begingroup$ This does not clear my question. First of all I don't want to combine $B$ and $C$, as I want to find their effect on $y$ too. Can I compare the effects of the matrices using the method I mentioned? Sorry for late reply, internet was cut off for 9 days at my region. $\endgroup$
    – impopularGuy
    Dec 20, 2019 at 5:29
  • $\begingroup$ If you want to replace $B$ by $B+C$, go right ahead. $$ \dfrac{d}{dt} \det(tA+B+C) = \text{tr}(\text{adj}(tA+B+C)A) =\det(tA+B+C)\; \text{tr}\left((tA+B+C)^{-1} A\right)$$ where for the last equation we assume $tA+B+C$ is invertible. $\endgroup$
    – Robert Israel
    Dec 20, 2019 at 13:13

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