I was looking for theorems that might be helpful in order for some proofs that I have and I came across the following one:

$$\frac{d}{dt} [\det A(t)]=\det A(t) \cdot \operatorname*{tr}[A^{-1}(t)\cdot \frac{d}{dt} A(t)]$$

where $A(t)$ is a matrix with a variable $t$.

The problem is that I have neither a reliable source for this theorem nor am I able to prove it.

Did anyone come across the aforementioned equation or is able to prove it?

  • 1
    $\begingroup$ Have you even tried searching for it? If I write "derivative determinant" on Google I am showered with relevant results, even on a fresh profile. $\endgroup$ Aug 17, 2015 at 8:42
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    $\begingroup$ This question really belongs to math.SE and I'm sure even there it's been asked a few times already! Voting to close. $\endgroup$
    – Suvrit
    Aug 17, 2015 at 12:42
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    $\begingroup$ Actually this question was indeed trivial and the answer was wikipedia-like, but I couldn't find any reference in my mother language and apparently googled the wrong words. I am sorry :/. $\endgroup$
    – Max
    Aug 17, 2015 at 21:33

2 Answers 2


This is just Jacobi's formula in the case of $A$ invertible.

Most books with any matrix theory in it should have a proof. Even wikipedia has one.


Another way to obtain the formula is to first consider the derivative of the determinant at the identity: $$ \frac{d}{dt} \det (I + t M) = \operatorname{tr} M. $$

Next, one has $$ \begin{split} \frac{d}{dt} \det A (t) &=\lim_{h \to 0} \frac{\det \bigl(A (t + h)\bigr) - \det A (t)}{h}\\ &=\det A (t) \lim_{h \to 0} \frac{\det \bigl(A (t)^{-1} A (t + h)\bigr) - 1}{h}\\ &=\det A (t) \operatorname{tr} \Bigl(A (t)^{-1}\frac{d A}{dt} (t) \Bigr). \end{split} $$

  • $\begingroup$ It’s essentially an application of the chain rule. $\endgroup$ Oct 10, 2022 at 16:33

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