Proof for the derivative of the determinant $d det(A)B = tr((cof(X))^{T}A)$ I want to proof that the following equations is true: $$d \space det(X)A = tr((cof(X))^{T}A)$$ I tried using Jacobi's formula, but I couldn't get anywhere. Has anyone an Idea? Kind Regards $cof(X)_{i, j} = (-1)^{i+j}*det(X)_{i, j}$ $(X)_{i, j}$ is equal to the Matrix $X$ where the $i^{th}$ row and $j^{th}$ column are cancelled

Let $X = (x_{i,j})_{i,j = 1}^n$ and let $X_{i,j}$ be the matrix obtained from $X$ by deleting the $i$-th row and $j$-th column. Then the expansion by minors formula is

$$ \det X = \sum_{j = 1}^n (-1)^{i + j} x_{i,j} \det X_{i,j}. $$

It follows that the partial derivative of $X \mapsto \det X$ with respect to $x_{i,j}$ is $(-1)^{i + j}\det X_{i,j}$ or more accurately, it is the map $A \mapsto \left( (-1)^{i + j}\det X_{i,j} \right) a_{i,j}$.

Hence the derivative at $X$ is the linear map

$$ A = (a_{i,j}) \mapsto \sum_{i,j = 1}^n \left. \frac{\partial \det}{\partial x_{i,j}} \right|_{X}(A) = \sum_{i,j = 1}^n \left( (-1)^{i + j} \det X_{i,j} \right)a_{i,j} = \operatorname{tr}\left( \operatorname{cof}(X)^T A \right). $$

So $(d\det)_XA = \operatorname{tr}( \operatorname{cof}(X)^T A )$.