## 每週問題 September 14, 2015

Let $A=[a_{ij}(t)]$ be an $n\times n$ matrix, where each entry $a_{ij}(t)$ is a differentiable function of $t$. Prove that

$\displaystyle \frac{d\det A}{dt}=\det B_1+\cdots+\det B_n$,

where $B_k$ is identical to $A$ except that the entries in the $k^{th}$ column are replaced by their derivatives, i.e., $(B_k)_{ij}=a_{ij}$ if $k\neq j$, $\displaystyle(B_k)_{ij}=\frac{da_{ij}}{dt}$ if $k=j$.

$\displaystyle \det A=\sum_{p}\sigma(p)a_{p_11}a_{p_22}\cdots a_{p_nn}$

\displaystyle \begin{aligned} \frac{d\det A}{dt}&=\frac{d}{dt}\sum_{p}\sigma(p)a_{p_11}a_{p_22}\cdots a_{p_nn}\\ &=\sum_p\sigma(p)\frac{d(a_{p_11}a_{p_22}\cdots a_{p_nn})}{dt}\\ &=\sum_p\sigma(p)\left(\frac{da_{p_11}}{dt}a_{p_22}\cdots a_{p_nn}+a_{p_11}\frac{da_{p_22}}{dt}\cdots a_{p_nn}+\cdots+a_{p_11}a_{p_22}\cdots \frac{da_{p_nn}}{dt}\right)\\ &=\sum_p\sigma(p)\frac{da_{p_11}}{dt}a_{p_22}\cdots a_{p_nn}+\sum_p\sigma(p)a_{p_11}\frac{da_{p_22}}{dt}\cdots a_{p_nn}+\cdots\\ &~~~~~+\sum_p\sigma(p)a_{p_11}a_{p_22}\cdots \frac{da_{p_nn}}{dt}\\ &=\det B_1+\det B_2+\cdots+\det B_n. \end{aligned}