How to find the trace of exponential of a matrix

A short route, which avoids unneeded diagonalizations, is to compute every exponential of $A$, that is, to look for functions $x$ and $y$ such that, for every $t$, $$e^{tA}=x(t)A+y(t)I$$ Then, $\mathrm{tr}(A)=-3$ and $\det(A)=2$ hence $$A^2=-3A-2I$$ This implies $$x'(t)A+y'(t)I=(e^{tA})'=Ae^{tA}=A(x(t)A+y(t)I)=x(t)(-3A-2I)+y(t)A$$ that is, $$x'(t)A+y'(t)I=(y(t)-3x(t))A-2x(t)I$$ or, equivalently, $$x'(t)=y(t)-3x(t)\quad y'(t)=-2x(t)$$ This implies that $$x''(t)+3x'(t)+2x(t)=0$$ The roots of the polynomial $t^2+3t+2=(t+2)(t+1)$ are $-1$ and $-2$ hence $$x(t)=ue^{-t}+ve^{-2t}$$ for some given constants $(u,v)$, which implies $$y(t)=x'(t)+3x(t)=2ue^{-t}+ve^{-2t}$$ The initial conditions $x(0)=0$ and $y(0)=1$, when applied to these formulas for $x(0)$ and $y(0)$, yield $u=1$ and $v=-1$, hence, for every $t$, $$e^{tA}=(e^{-t}-e^{-2t})A+(2e^{-t}-e^{-2t})I$$ that is, $$e^{tA}=\begin{pmatrix}-e^{-t}+2e^{-2t} & 2e^{-t}-2e^{-2t}\\-e^{-t}+e^{-2t}&2e^{-t}-e^{-2t}\end{pmatrix}$$ from which the value of the matrix $e^{3A}$ follows.