Let $A:[0,T]\to \mathbb{R}^{n\times n}$ be a continuous function, and let $\Phi$ satisfy $\frac{d}{d t} \Phi_t=A_t \Phi_t$ for all $t\in [0,T]$, and $\Phi_0=I_n$. I hope to prove that the solution to the linear equation $$ \frac{d}{d t} \Psi_t=- \Psi_tA_t, \quad t\in [0,T]; \quad \Psi_0=I_n, \tag1 $$ is the matrix inverse of $\Phi$, i.e. ,$\Psi_t:=\Phi_t^{-1}$ for all $t\in [0,T]$. By definition, $\Psi_0\Phi_0=\Phi_0\Psi_0=I$. Hence it suffices to verify $\frac{d}{dt}(\Psi\Phi)_t=\frac{d}{dt}(\Phi\Psi)_t=0$ for all $t\in [0,T]$. The identity $\frac{d}{dt}(\Psi\Phi)_t=0$ is easy, but I met some difficulty in proving $\frac{d}{dt}(\Phi\Psi)_t=0$. Below please find my calculation: \begin{align*} \frac{d}{dt}(\Phi_t\Psi_t)^i_j=\sum_{k}\frac{d}{dt}\left((\Phi_t)^i_k(\Psi_t)^k_j\right) =\sum_{k,l}\left((A_t)^i_l(\Phi_t)^l_k(\Psi_t)^k_j+(\Phi_t)^i_k\left(-(\Psi_t)^k_l(A_t)^l_j\right)\right), \end{align*} which does not seem to be zero. Have I overlooked anything?
Note that $\Phi_t$ is invertible and $\Psi_t$ is a left-inverse of $\Phi_t$. Hence by the invertible matrix theorem, $\Psi_t$ should be the unique right-inverse of $\Phi_t$.