Show that: a) $X^{-1}(t)$ is bounded in $[\beta,\infty)$. b)No system solution approaches zero solution when $t \rightarrow \infty.$ Let a system $x' = A(t)x$ and suppose there are values positives $k, \beta$ such that a positive fundamental matrix $X(t)$ satisfies
$\|X(t)\| \leq k$, $t \geq \beta$ and
$$ \liminf_{t \rightarrow \infty} \int^t_\beta \operatorname{tr}(A(s))\,ds > - \infty.$$
Show that:
a)$X^{-1}(t)$ is bounded in $[\beta,\infty)$.
b)No system solution approaches zero solution when $t \rightarrow \infty.$
 A: Here is a partial answer.  Assume $X(t)$ and $A(t)$ are $n \times n$ matrices. Define: 
$$ c = \liminf_{t\rightarrow\infty} \int_\beta^t tr(A(s))ds $$ 
The problem tells us that $c > -\infty$. 
1) Note that since $X(t)$ is a solution to $X'(t) = A(t)X(t)$, then for any $n\times n$ matrix $M$ we also get $Y(t) = X(t)M$ is another solution. Let $y_{\beta}$ be any desired initial condition for $Y(\beta)$.  If $X(\beta)$ is invertible, then we can choose $M = X(\beta)^{-1}y_{\beta}$, so that $Y(t)=X(t)M$ is a solution with the desired initial condition $Y(\beta)=y_{\beta}$. 
Thus, limiting behavior about $X(t)$ can often be translated to limiting behavior about other solutions $Y(t)$. If $Y(\beta)$ is invertible and $X(t)$ does not approach the zero solution, then $Y(t)$ also does not approach the zero solution. (If $Y(\beta)$ is not invertible then we can have $Y(t)=0$ for all $t$ as a valid solution). 
2) Suppose $A(t)$ is a matrix of the form $A(t) = g(t)A$ for some (constant) matrix $A$ and some scalar-valued integrable function $g(t)$.  Then a solution to the ODE is (for $t \geq \beta$):
$$ X(t) = e^{\int_{\beta}^t A(s)ds}X(\beta) \: \: (Equation 1) $$
There are other types of $A(t)$ matrices for which the above solution works, but not all (as pointed out by helpful comments of Robert Lewis above). 
Now, the following link gives Jacobi's formula for any square matrix $B$: $\det(e^B) = e^{tr(B)}$. 
http://en.wikipedia.org/wiki/Matrix_exponential
Applying this to (Equation 1) gives: 
$$ \det(X(t)) = e^{\int_\beta^ttr(A(s))ds}\det(X(\beta))$$
Taking absolute values of both sides gives: 
$$ |\det(X(t))| = e^{\int_{\beta}^t tr(A(s))ds} |\det(X(\beta))| $$
Taking a $\liminf$ of both sides gives: 
$$ \liminf_{t\rightarrow\infty} |\det(X(t))| = e^{c} |\det(X(\beta))| > 0 $$
where the final inequality holds because we are told that $\det(X(\beta))\neq 0$.  Since the absolute value of the determinant is bounded away from $0$, and since the determinant will be a polynomial function of the entries of the matrix, it follows that the sum of squares of the entries of $X(t)$ must be bounded away from zero. 
A: A quick comment to @Michael's answer: I guess we can apply Liouville's Formula here, which only requires $tr(A(t))$ to be a continuous function. Applying Liouville's Formula gives
$$det(X(t))=det(X(\beta))e^{\int_{\beta}^t tr(A(s))ds},$$
and the rest is given in @Michael 's answer:) 
