In the $2\times2$ case, you can find the exponential of a matrix $A$ without having to decompose it into $BMB^{-1}$ form. In particular, you only need the eigenvalues—you don’t need to find any eigenvectors. There are three cases, as follows.
Distinct Real Eigenvalues: Let $P_1 = (A-\lambda_2I)/(\lambda_1-\lambda_2)$ and $P_2 = (A-\lambda_1I)/(\lambda_2-\lambda_1)$, where $\lambda_1,\lambda_2$ are the eigenvalues. These two matrices are projections onto the eigenspaces corresponding to $\lambda_1$ and $\lambda_2$, respectively. They have some handy properties: $P_1P_2=P_2P_1=0$, $P_1+P_2=I$, and $A=\lambda_1P_1+\lambda_2P_2$. Using these projections, $$\exp(tA)=\exp(t\lambda_1P_1+t\lambda_2P_2)=\mathrm e^{\lambda_1t}P_1+\mathrm e^{\lambda_2t}P_2.$$
Repeated Eigenvalue: Let $G=A-\lambda I$, where $\lambda$ is the eigenvalue. By the Cayley-Hamilton theorem, $(A-\lambda I)^2=0$, so $G$ is nilpotent. Since $G$ and $\lambda I$ commute, $\exp(tA)=\exp(\lambda tI)\exp(tG)$, but $\exp(tG)=I+tG$ (expand using the power series), so $$\exp(tA) = \mathrm e^{\lambda t}(I+tG).$$
Complex Eigenvalues: The eigenvalues are of the form $\lambda=\alpha\pm\mathrm i\beta$, and the characteristic equation is $(\lambda-\alpha)^2+\beta^2=0$. By the Cayley-Hamilton theorem, $(A-\alpha I)^2+\beta^2I=0$, so the traceless matrix $G=A-\alpha I$ satisfies $G^2=-\beta^2 I$. Writing $A=\alpha I+G$, we have $\exp(tA)=\exp(\alpha I)\exp(tG)$. Expanding the latter as a power series and using the above equality, $$
\exp(tG) = I+tG-\frac{\beta^2t^2}{2!}I-\frac{\beta^2t^3}{3!}G+\frac{\beta^4t^4}{4!}I+\dots.
$$ The coefficient of $I$ is the power series for $\cos{\beta t}$, while the coefficient of $G$ is the power series for $(\sin{\beta t})/\beta$. Thus, $$
\exp(tG) = \cos{\beta t}\;I+[(\sin{\beta t})/\beta]\;G \\
\text{and} \\
\exp(tA) = \mathrm e^{\alpha t}\exp(tG).
$$
