= Solution
The upper-triangular <matrix> has distinct <eigenvalues> $2,-1$, with respective <eigenvectors> $(1,0)^T$ and $(-2,1)^T$. Thus
$$
A=S\begin{pmatrix}2&0\\0&-1\end{pmatrix}S^{-1},\qquad
S=\begin{pmatrix}1&-2\\0&1\end{pmatrix},\qquad
S^{-1}=\begin{pmatrix}1&2\\0&1\end{pmatrix}.
$$
The defining <power series> shows that the <matrix exponential respects similarity>. Applying <11/a/solution> to the diagonal factor gives
$$
\boxed{e^A=S\begin{pmatrix}e^2&0\\0&e^{-1}\end{pmatrix}S^{-1}
=\begin{pmatrix}e^2&2(e^2-e^{-1})\\0&e^{-1}\end{pmatrix}.}
$$
As a direct series check, $(A^m)_{12}=2(2^m-(-1)^m)$ for $m\geq0$; summing this entry divided by $m!$ gives the same off-diagonal expression.
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