行列式の微分

$ \mathrm d\det\bm A=(\det\bm A){{\bm A}^\top}^{-1}:\mathrm d\bm A

2次元のとき

$ \mathrm d\det\bm A=\mathrm d(A_{00}A_{11}-A_{01}A_{10})

$ =(\mathrm d A_{00})A_{11}+A_{00}\mathrm dA_{11}-(\mathrm d A_{01})A_{10}-A_{01}\mathrm dA_{10}

$ = \begin{pmatrix}A_{11}&-A_{10}\\-A_{01}&A_{00}\end{pmatrix}:\mathrm d\begin{pmatrix}A_{00}&A_{01}\\A_{10}&A_{11}\end{pmatrix}

$ = (\det\bm A){{\bm A}^\top}^{-1}:\mathrm d\bm A

一般の導出

固有値を使った方法

$ \mathrm d\det\bm A=\det(\bm A+\mathrm d\bm A)-\det\bm A

$ =\det(\bm A\cdot(\bm I+{\bm A}^{-1}\cdot\mathrm d\bm A))-\det\bm A

$ = (\det\bm A)(\det(\bm I+\bm A^{-1}\cdot\mathrm d\bm A)-1)

$ = (\det\bm A)(\det(\bm I+\bm A^{-1}\cdot(\bm 0+\mathrm d\bm A))-\det(\bm I+\bm A^{-1}\cdot\bm0))

$ = (\det\bm A)\left.\frac{\mathrm d}{\mathrm d\epsilon}\det(\bm I+\bm A^{-1}\cdot\epsilon d\bm A)\right|_{\epsilon=0}

$ = (\det\bm A)\left.\frac{\mathrm d}{\mathrm d\epsilon}\prod_i\left(\epsilon\lambda_i^{{\bm A}^{-1}\cdot\mathrm d\bm A}-(-1)\right)\right|_{\epsilon=0}

$ = (\det\bm A)\left.\sum_j\lambda_j^{{\bm A}^{-1}\cdot\mathrm d\bm A}\prod_{i\neq j}\left(1+\epsilon\lambda_i^{{\bm A}^{-1}\cdot\mathrm d\bm A}\right)\right|_{\epsilon=0}

$ = (\det\bm A)\sum_j\lambda_j^{{\bm A}^{-1}\cdot\mathrm d\bm A}

$ = (\det\bm A)\mathrm{tr}\left({\bm A}^{-1}\cdot\mathrm d\bm A\right)

$ \underline{=(\det\bm A){{\bm A}^\top}^{-1}:\mathrm d\bm A\quad}_\blacksquare