How to show that the solution matrix in a matrix differential equation has nonzero determinant I have the following problem:
Consider the matrix differential equation
$$\frac{d}{dt}X(t)=A(t)X(t)$$
$$X(t_0)=X_0$$
Show that if $\det(X_0)\neq0$, then $\det(X(t))\neq0$ for all $t\ge t_0$.
I've been thinking about this for awhile now but have yet to make any progress so I'm wondering if I'm missing an obvious proof, or if it's actually not very trivial. Any help is appreciated. Thanks!
 A: The crucial property for your result is

$$\frac{d}{dt}\left[\det X(t)\right]=tr[A(t)]\det[X(t)]\qquad
 \qquad\qquad(1)$$

from which one directly obtains
$$\det X(t)=\exp\left(\int_{t_0}^t{tr[A(s)]ds}\right)\det(X_0)$$
Thus if $\det(X_0)\neq 0$ then $\det(X(t))\neq 0$ for all $t\geq t_0$.

Proof of (1):
Let $c_{ij}(t)$ be the cofactor of the entry $X_{ij}(t)$ of matrix $X(t)$. If $Ad(t)$ is the adjoint matrix of $X(t)$ then 
$$ X(t) Ad(t) =\det(X(t))\mathbb{I}_n.$$
Also it is well known that the $(i,j)$ element $Ad_{ij}(t)$ of $Ad(t)$ is $c_{ji}(t)$ i.e. the adjoint is the transpose of the cofactor matrix.  
For the time derivative of $\det X(t)$ we have
$$\frac{d}{dt}[\det X(t)] =\sum_{i=1}^n\sum_{j=1}^n{\frac{\partial \det X(t)}{\partial X_{ij}(t)}\frac{d}{dt}[X_{ij}(t)]}$$
Also by definition
$$\det X(t)=\sum_{i=1}^n{c_{ij}(t)X_{ij}(t)}$$
and therefore 
$$\frac{\partial \det X(t)}{\partial X_{ij}(t)}=c_{ij}(t)=Ad_{ji}(t)$$
Thus
$$\frac{d}{dt}[\det X(t)] =\sum_{i=1}^n\sum_{j=1}^n{Ad_{ji}(t)\frac{d}{dt}[X_{ij}(t)]}\\
=tr\left[Ad(t)\frac{dX}{dt}\right]=tr\left[Ad(t)A(t)X(t)\right]\\
=tr\left[(X(t)Ad(t))\:A(t)\right]=tr\left[\det(X(t))A(t)\right]=tr\left[A(t)\right]\det(X(t))$$
where I have used the property $tr(AB)=tr(BA)=\sum_{i=1}^n\sum_{j=1}^n{a_{ij}b_{ji}}$ for square matrices $A=[a_{ij}], B=[b_{ij}]$.
