= Solution
For distinct <eigenvalues>, the formula holds for $n=0,1$. If the lower-left entry of $L^n$ is $b_n=(\lambda_1^n-\lambda_2^n)/(\lambda_1-\lambda_2)$, multiplication by $L$ gives $b_{n+1}=\lambda_1b_n+\lambda_2^n=(\lambda_1^{n+1}-\lambda_2^{n+1})/(\lambda_1-\lambda_2)$. This proves by <mathematical induction> that
$$
L^n=\begin{pmatrix}\lambda_1^n&0\\(\lambda_1^n-\lambda_2^n)/(\lambda_1-\lambda_2)&\lambda_2^n\end{pmatrix}.
$$
Summing the absolutely convergent <matrix exponential> series gives
$$
\boxed{A(t)=e^{Lt}=\begin{pmatrix}e^{\lambda_1t}&0\\\dfrac{e^{\lambda_1t}-e^{\lambda_2t}}{\lambda_1-\lambda_2}&e^{\lambda_2t}\end{pmatrix}.}
$$
For the repeated <eigenvalue> $\lambda_1=\lambda_2=\lambda$, the continuous limit is $A(t)=e^{\lambda t}\begin{pmatrix}1&0\\t&1\end{pmatrix}$. This is also the exponential of a two-dimensional <Jordan block>.
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