# Proof for the derivative of the determinant of a matrix [closed]

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

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

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?

## closed as off-topic by Suvrit, Boris Bukh, Alex Degtyarev, Chris Godsil, Will SawinAug 17 '15 at 14:17

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• 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. – Federico Poloni Aug 17 '15 at 8:42
• This question really belongs to math.SE and I'm sure even there it's been asked a few times already! Voting to close. – Suvrit Aug 17 '15 at 12:42
• 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 :/. – Max Aug 17 '15 at 21:33

This is just Jacobi's formula in the case of $A$ invertible.
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 $$\begin{split} \frac{d}{dt} \det A (t) &=\lim_{h \to 0} \frac{\det \bigl(A (t + h)\bigr) - \det A (t)}{h}\\ &=\det A (t) \lim_{h \to 0} \frac{\det \bigl(A (t)^{-1} A (t + h)\bigr) - 1}{h}\\ &=\det A (t) \operatorname{tr} \Bigl(A (t)^{-1}\frac{d A}{dt} (t) \Bigr). \end{split}$$