Show that if $\lim_{t\to\infty} \int_0^t \operatorname{tr}(A(s))ds=\infty$, then $x'(t) = A(t)x(t)$ has an unbounded solution. Show that if $\lim_{t\to\infty} \inf \int_0^t \operatorname{tr}(A(s))ds = \infty$, then $x'(t) = A(t)x(t)$ has an unbounded solution. (Hint: Use $W (t) = W (t_0) \exp \left( \int_{t_0}^t \operatorname{tr}(A(s))ds \right)$ Abel’s identity or Liouville’s formula.)
The matrix $\Pi(t, t_0)$ is the principal matrix solution and it solves
$\Pi'(t, t_0) = A(t)\Pi(t, t_0), \;\;\;\;\;\;\; \Pi(t_0, t_0) = I$
But but how to achieve a relationship with the Wronskian to solve this problem?
Would it be right to do that?
$\det(\Pi(t, 0)) = \det(\Pi(0, 0))\exp \left(\int_{0}^t \operatorname{tr}(A(s))ds \right) \Rightarrow \det(\Pi(t, 0)) = \det(I)\cdot \exp \left( \int_{0}^t \operatorname{tr}(A(s))ds \right)
\Rightarrow \det(\Pi(t, 0)) = 1\cdot\exp \left(\int_{0}^t \operatorname{tr}(A(s))ds \right) 
\Rightarrow \ln(\det(\Pi(t, 0))) = \int_{0}^t \operatorname{tr}(A(s))ds 
\Rightarrow \lim_{t\to\infty}\inf(\ln(\det(\Pi(t, 0)))) = \lim_{t\to\infty} \inf \int_0^t \operatorname{tr}(A(s))ds = \infty$
and then conclude that some of the solutions of $\Pi'(t, t_0)$ tend to infinity?
 A: Our OP is mos' def' on the right track, in fact has basically answered his own question, I'll warrant.
These things being said, here's my write-up on this one:
The most straightforward solution I know is to invoke Liouville's Formula, which in general form states that if $Y(t)$ is a square matrix solution to 
$\dot Y(t) = A(t) Y(t), \tag 1$
then $\det Y(t)$ is given by
$\det Y(t) = \det Y(t_0) \exp \left ( \displaystyle \int_{t_0}^t \text{Tr}(A(s)) \; ds \right ), \tag 2$
where if
$A(t) = [a_{ij}(t)], \tag 3$
we have
$\text{Tr}(A(t)) = \displaystyle \sum_1^n a_{ii}(t), \tag 4$
i.e., $\text{Tr}(A(t))$ is the sum of the diagonal entries of $A(t)$.  If we now take
$Y(t_0) = I, \tag 5$
we find that 
$\det Y(t) = \exp \left ( \displaystyle \int_{t_0}^t \text{Tr}(A(s)) \; ds \right ). \tag 6$
Now if $\vec b$ is any (fixed) column vector, then
$(Y(t)\vec b)' = \dot Y(t) \vec b = A(Y(t) \vec b), \tag 7$
which shows that $Y(t) \vec b$ is a column-vector solution to 
$\dot x(t) = A(t)x(t); \tag{8}$
clearly,
$Y(t_0) \vec b = \vec b, \tag 9$
the initial condition for (7) at $t_0$.  It follows then that every solution to (8) is a linear combination of the columns of $Y(t)$ with fixed (that is, time-invariant) coefficients which form the vector $\vec b$.
Now if every solution to (8) is bounded, so is every column of $Y(t)$, since they are individually solutions to (8); this in turn implies that the individual components of each column of $Y(t)$ are also bounded; since then all the entries of $Y(t)$ are bounded we conclude that $\det Y(t)$ is bounded itself; that is, there exists some real $M > 0$ with
$ \exp \left ( \displaystyle \int_{t_0}^t \text{Tr}(A(s)) \; ds \right ) = \det Y(t) < M, \; \forall t \in \Bbb R; \tag{10}$
we note that since $\exp ( \int_{t_0}^{t_0} \text{Tr}(A(s)) \; ds ) = 1$, we have $M > 1$, so $\ln M > 0$ and we may write
$ \displaystyle \int_{t_0}^t \text{Tr}(A(s)) \; ds < \ln M, \; \forall t \in \Bbb R; \tag{11}$
then
$\displaystyle \lim_{t \to \infty} \inf \int_{t_0}^t \text{Tr}(A(s)) \; ds = \lim_{t \to \infty} (\inf_{u \ge t} \{\int_{t_0}^u \text{Tr}(A(s)) \; ds \})$ $\le \lim_{t \to \infty} (\inf_{u \ge t} \{\ln M \}) = \lim_{t \to \infty}  ( \ln M ) = \ln M < \infty, \tag{12}$
which contradicts the hypothesis that 
$\displaystyle \lim_{t \to \infty} \inf \int_{t_0}^t \text{Tr}(A(s)) \; ds = \infty; \tag{13}$
thus some column of $Y(t)$, that is, some solution of (8), must become unbounded as $t \to \infty$.
