If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, then $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$ 
If $A$ is a $2\times 2$ matrix with a repeated eigenvalue $r$, show that  $\mathrm{e}^{At}=\mathrm{e}^{rt}\left[I+(A-rI)t\right]$.

I have already been able to show that if $A$ is an arbitrary $2\times 2$ matrix 
$$
\left(
\begin{array}\\
a & b \\
c & d
\end{array}
\right),
$$
the repeated eigenvalue $r$ must be equal to
$\,r=\dfrac{a+d}{2}$.
From here, my first thought was to take the determinant of the matrix
$$\mathrm{e}^{rt}\left[I+(A-rI)t\right],$$ 
and show that it is equal to $$\det(\mathrm{e}^{At}),$$ but then I started thinking this would be useless since two unequal matrices can have the same determinant.
I've tried directly substituting $r$ into the right-hand side of the equation as well, but with no luck.

Does anyone have any suggestions on how to proceed from here?

As $r$ is an eigenvalue of multiplicity 2 of the $2\times 2$ matrix $A$, 
then $A$ does not have any other eigenvalues, and its characteristic polynomial should be
$$
p(x)=(x-r)^2,
$$
and by Cayley-Hamilton Theorem we have that
$(A-rI)^2=0$.
Next observe that
$$
\mathrm{e}^{tA}=\mathrm{e}^{rt}\mathrm{e}^{t{(A-rI)}}=\mathrm{e}^{rt}\sum_{n=0}^\infty
\frac{(A-rI)^n}{n!}=\mathrm{e}^{rt}\sum_{n=0}^1
\frac{(A-rI)^n}{n!}=\mathrm{e}^{rt}\big(I+
(A-rI)\big)
$$
