Exponential of matrix So, I'm wondering if there is an easy way (as in not calculating the eigenvalues, Jordan canonical form, change of basis matrix, etc.) to calculate this exponential $e^{At}$
with $$A=\begin{pmatrix} 0&9 \\ -1&0 \end{pmatrix}.$$
I'd suppose it stumbles around cosines and sines, but I'm not really sure... In fact, something of the kind 
$$A=\begin{pmatrix} 3 & 5\cr-5 & -3\end{pmatrix}$$
would also interest me.
 A: Let
$$A=\pmatrix{0&9\cr-1&0\cr}\ .$$
You can easily check that
$$A^2=-9I$$
and so the exponential series gives
$$\eqalign{e^{tA}
  &=I+tA+\frac{1}{2!}t^2A^2+\frac{1}{3!}t^3A^3+\cdots\cr
  &=\Bigl(I-\frac{1}{2!}(9t^2I)+\frac{1}{4!}(9^2t^4I)+\cdots\Bigr)+\Bigl(tA-\frac{1}{3!}(9t^3A)+\frac{1}{5!}(9^2t^5A)+\cdots\Bigr)\cr
  &=(\cos3t)I+\Bigl(\frac{\sin3t}{3}\Bigr)A\ .\cr}$$
For your other matrix you get $B^2=-16I$ and the same method works.
A: If one is allowed to use the values $a$ and $b$ of the eigenvalues, and if $a\ne b$, then one gets 

$$\mathrm e^{tA}=\frac{\mathrm e^{at}-\mathrm e^{bt}}{a-b}A+\frac{a\mathrm e^{bt}-b\mathrm e^{at}}{a-b}I,$$ 

No need here for series expansion or "Jordan canonical form, change of basis matrix, etc.", the substitute being to work directly in the two-dimensional vector space $\mathbb C[A]$, with basis $\{I,A\}$.

To prove the formula above, consider the function $$u(t)=\mathrm e^{At}.$$ The facts that $u'(t)=A\mathrm e^{At}$, $u''(t)=A^2\mathrm e^{At}$, and $A^2-(a+b)A+ab=0$, together yield $$u''(t)-(a+b)u'(t)+abu(t)=0.$$ This (matrix-valued) differential equation is solved by $$u(t)=\mathrm e^{at}U+\mathrm e^{bt}V,$$ for some well chosen (matrices) $U$ and $V$. These are such that $$U+V=u(0)=I,\qquad aU+bV=u'(0)=A.$$ Solving this $2\times2$ Cramer system for $(U,V)$ yields the formula stated at the beginning of this post. 
In the end, the method only requires the (seemingly inescapable) step to solve for $(a,b)$ the system 

$$a+b=\mathrm{tr}(A),\qquad ab=\mathrm{det}(A).$$

Example 1: If $A=\begin{pmatrix} 0&9 \\ -1&0 \end{pmatrix}$, then $a+b=0$ and $ab=9$ hence one can choose $a=3\mathrm i$ and $b=-3\mathrm i$, then $a-b=6\mathrm i$, $\mathrm e^{at}-\mathrm e^{bt}=2\mathrm i\sin(3t)$ and $a\mathrm e^{bt}-b\mathrm e^{at}=6\mathrm i\cos(3t)$ hence
$$\mathrm e^{tA}=\tfrac13\sin(3t)A+\cos(3t)I=\begin{pmatrix} \cos(3t)&3\sin(3t) \\ -\tfrac13\sin(3t)&0\cos(3t)\end{pmatrix}.$$
Example 2: If $A=\begin{pmatrix} 3 & 5\cr-5 & -3\end{pmatrix}$, then $a+b=0$ and $ab=16$ hence one can choose $a=4\mathrm i$ and $b=-4\mathrm i$, then $a-b=8\mathrm i$, $\mathrm e^{at}-\mathrm e^{bt}=2\mathrm i\sin(4t)$ and $a\mathrm e^{bt}-b\mathrm e^{at}=8\mathrm i\cos(4t)$ hence
$$\mathrm e^{tA}=\tfrac14\sin(4t)A+\cos(4t)I=\begin{pmatrix} \cos(4t)+\tfrac34\sin(4t)&\tfrac54\sin(4t) \\ -\tfrac54\sin(4t)&\cos(4t)-\tfrac34\sin(4t)\end{pmatrix}.$$

Edit: As noted by @abel in a comment, Putzer algorithm can be used to compute $\mathrm e^{tA}$. For $A$ of size $2\times2$ with eigenvalues $a$ and $b$, the algorithm says that $$\mathrm e^{tA}=r(t)I+s(t)(A-aI),$$ where the (scalar) functions $r$ and $s$ solve the differential system $$r'(t)=ar(t),\quad r(0)=1,\quad s'(t)=bs(t)+r(t),\quad s(0)=0.$$ This yields $r(t)=\mathrm e^{at}$ and, since $a\ne b$, $(a-b)s(t)=\mathrm e^{at}-\mathrm e^{bt}$, thus $$\mathrm e^{tA}=\mathrm e^{at}I+\frac{\mathrm e^{at}-\mathrm e^{bt}}{a-b}\,(A-aI),$$ which is equivalent to the formula we proved above.
A: These two matrices have a null trace and a positive determinant, so that their Eigenvalues are pure imaginary conjugate. You can infer that the solution will be of the form $C\cos\lambda t+S\sin\lambda t$ (real linear combinations of imaginary exponentials), where $\lambda=\sqrt\Delta$.
Now for $t=0$,
$$C\cos\lambda0+S\sin\lambda0=C=e^{A0}=I,$$
and evaluating the first derivative at $t=0$,
$$-\lambda C\sin\lambda0+\lambda S\cos\lambda0=\lambda S=Ae^{A0}=A.$$
Thus the final solution
$$I\cos\lambda t+\frac A\lambda\sin\lambda t.$$
A: We can always find the exponential of a $2 \times 2$  matrix ( with real or complex entries) without calculating eigenvalues or Jordan form.
1) Note that we can always put a $2 \times 2$ matrix in the form $A=hI+A'$ where: $h=\mbox{tr}A/2$, $\mbox{tr}A'=0$, $\mbox{det}A'=\mbox{det}A-\left(\dfrac{\mbox{tr}A}{2}\right)^2$; and $[hI,A']=0$ so that $e^A=e^he^{A'}$.
2) Note that for a traceless matrix $M$ we have:
$$
M^2=\mbox{det}(M)I
$$
so that, given $\theta=\sqrt{\mbox{det}(M)}$, we have:
$$
\begin{split}
e^M&=\sum_{k=0}^\infty\dfrac{M^k}{k!}=\\
&=I+\dfrac{M}{1!}-\dfrac{\theta^2I}{2!}-\dfrac{\theta^3M}{3!\,\theta}+\dfrac{\theta^4I}{4!}+\dfrac{\theta^5M}{5!\,\theta}-\dfrac{\theta^6I}{6!}+\cdots=\\
&=I \left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\dfrac{\theta^6}{6!}\cdots\right)+\dfrac{M}{\theta}\left( \dfrac{\theta}{1!}-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}\cdots\right)=\\
&=I\cos\theta +\dfrac{M}{\theta}\sin\theta
\end{split}
$$
3) So: for every   $2 \times 2$ matrix  $A=hI+A'$ we have
$$
e^A=e^h\left(I\cos\theta +\dfrac{A'}{\theta}\sin\theta\right) \qquad \theta=\sqrt{\mbox{det}(A')}
$$
