Derivative of the Wronskian Consider a non-autonomous linear system of ode's: $X'=A(t)X$, $X:\mathbb{R}\rightarrow\mathbb{R}^n$. Let $B(t)$ be a fundamental matrix solution $\dot{B}=A(t)B$ of the system and $W(t):=\det B(t)$ the Wronskian. Show that
$$\dot{W}=tr(A(t))W.$$
My idea: I started by induction. For $n=2$ we have $B=\left(\begin{array}{cc}
x_{1} & y_{1}\\
x_{2} & y_{2}
\end{array}\right)$ and
$$W =\det B =x_{1}y_{2}-y_{1}x_{2}.$$
Therefore,
$$
\begin{eqnarray}
\dot{W} &=&\dot{x_{1}}y_{2}+x_{1}\dot{y_{2}}-\dot{y_{1}}x_{2}-y_{1}\dot{x_{2}}\\
 &=&\dot{x_{1}}y_{2}-\dot{y_{1}}x_{2}+x_{1}\dot{y_{2}}-y_{1}\dot{x_{2}}\\
 &=&\det\left(\begin{array}{cc}
\dot{x_{1}} & x_{2}\\
\dot{y_{1}} & y_{2}
\end{array}\right)+\det\left(\begin{array}{cc}
x_{1} & \dot{x_{2}}\\
y_{1} & \dot{y_{2}}
\end{array}\right)\\
 &=&\det\left(\begin{array}{cc}
a_{11}x_{1}+a_{12}x_{2} & x_{2}\\
a_{11}y_{1}+a_{12}y_{2} & y_{2}
\end{array}\right)+\det\left(\begin{array}{cc}
x_{1} & a_{21}x_{1}+a_{22}x_{2}\\
y_{1} & a_{21}y_{1}+a_{22}y_{2}
\end{array}\right)\\
 &=&\det\left(\begin{array}{cc}
a_{11}x_{1} & x_{2}\\
a_{11}y_{1} & y_{2}
\end{array}\right)+\det\left(\begin{array}{cc}
a_{12}x_{2} & x_{2}\\
a_{12}y_{2} & y_{2}
\end{array}\right)+\det\left(\begin{array}{cc}
x_{1} & a_{21}x_{1}\\
y_{1} & a_{21}y_{1}
\end{array}\right)+\det\left(\begin{array}{cc}
x_{1} & a_{22}x_{2}\\
y_{1} & a_{22}y_{2}
\end{array}\right)\\
 &=&\det\left(\begin{array}{cc}
a_{11}x_{1} & x_{2}\\
a_{11}y_{1} & y_{2}
\end{array}\right)+\det\left(\begin{array}{cc}
x_{1} & a_{22}x_{2}\\
y_{1} & a_{22}y_{2}
\end{array}\right)\\
 &=&a_{11}\det\left(\begin{array}{cc}
x_{1} & x_{2}\\
y_{1} & y_{2}
\end{array}\right)+a_{22}\det\left(\begin{array}{cc}
x_{1} & x_{2}\\
y_{1} & y_{2}
\end{array}\right)\\
 &=&(a_{11}+a_{22})\det\left(\begin{array}{cc}
x_{1} & x_{2}\\
y_{1} & y_{2}
\end{array}\right)\\
 &=&tr(A)W.
\end{eqnarray}$$
Now I suppose that it is valid for $n>2$ and consider taking $n+1$ and the formula for the determinant
$$W=\sum_{k=1}^{n+1}(-1)^{i+k}b_{ik}\det B_{ik}$$
where $B_{ik}$ is the $ik$-minor of $B$. Up to here I haven't been able to advance. Any suggestions?
 A: The link in the comments contains interesting responses towards solving the question, but here I would like to add a derivation based on the properties of the Levi-Civita symbol, which is very concise in my opinion, at the cost of handling some indices and tensor calculus conventions.
The determinant of the $n\times n$ matrix can be written
$$n!\det B=\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}B_{i_1j_1}...B_{i_nj_n}$$
where all the indices here are summed over their range. Since $\dot{B}_{i_kj_k}=A_{j_k \ell}B_{i_k\ell}$ we can take the derivative of this relation:
$$n!\dot{W}=\sum_k\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}A_{j_k \ell}B_{i_k\ell}\prod _{m\neq k}B_{i_mj_m}$$
Now notice that $\epsilon_{i_1...i_n}B_{i_1j_1}...B_{i_k\ell}...B_{i_nj_n}=\epsilon_{j_1...\ell...j_n}W $
which leads us, after reordering indices appropriately and taking account of the minus signs generated, to the relation
$$n!\dot{W}=W\sum_k\epsilon_{\ell j_1...j_n}\epsilon_{j_k j_1...j_n}A_{j_k \ell}$$
Here, note that $\epsilon_{\ell j_1...j_n}\epsilon_{j_k j_1...j_n}=(n-1)!\delta_{\ell j_k}$ which allows us to write
$$n!\dot{W}=(n-1)!W\sum_k\delta_{ \ell j_k }A_{j_k\ell}=(n-1)!\sum_{k=1}^n(\text{tr} A) W\Rightarrow\dot{W}=(\text{tr}A(t))W  $$
