How to find exponential of triangular matrix

Let $I = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$.

Since $A = -I+B$ and the matrices $I$ and $B$ commute, we have $e^{tA} = e^{t(-I+B)} = e^{-tI}e^{tB}$.

Trivially, $e^{-tI} = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-t}\end{bmatrix}$. Also, since $B^2 = 0$, we have $e^{tB} = I+tB = \begin{bmatrix} 1 & t \\ 0 & 1\end{bmatrix}$.

Therefore, $e^{tA} = e^{-tI}e^{tB} = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-t}\end{bmatrix}\begin{bmatrix} 1 & t \\ 0 & 1\end{bmatrix} = \begin{bmatrix} e^{-t} & te^{-t} \\ 0 & e^{-t}\end{bmatrix}$.

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Better answer: Suppose $\begin{bmatrix} e^{-t} & te^{-t} \\ 0 & e^{-t}\end{bmatrix} = e^{tA}$ for some matrix $A$.

Differentiate to get $\begin{bmatrix} -e^{-t} & (1-t)e^{-t} \\ 0 & -e^{-t}\end{bmatrix} = Ae^{tA}$. Then plug in $t = 0$ to get $\begin{bmatrix} -1 & 1 \\ 0 & -1\end{bmatrix} = A$.

This assumes that the given matrix is the matrix exponential of some matrix, but if it is a true/false question and you don't have a lot of time, making that assumption might be the way to go.