Using the Sylvester interpolation formula to calculate $e^{A t}$, what is the result?
$$
\mathbf{A}=\left[ \begin{matrix} {{2}} & {{1}} & {{4}} \\ {{0}} & {{2}} & {{0}} \\ {{0}} & {{3}} & {{1}} \\ \end{matrix} \right]
$$

Options:
A. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{t}}}} & {{{1 2 \mathrm{e}^{t} \,-\, t \mathrm{e}^{t} \,+\, \mathrm{e}^{2 t}}}} & {{{-4 \mathrm{e}^{t} \,+\, 3 \mathrm{e}^{t}}}} \\ {{{0}}} & {{{\mathrm{e}^{2 t}}}} & {{{3 t \mathrm{e}^{t}}}} \\ {{{1}}} & {{{\mathrm{e}^{t} \,-\, 3}}} & {{{\mathrm{e}^{2 t}}}} \\ \end{array} \right]
$$
B. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{3 t}}}} & {{{1 2 \mathrm{e}^{t} \,+\, 1 5 t \,+\, 1 3 t \mathrm{e}^{t}}}} & {{{-4 \mathrm{e}^{t} \,+\, 2 \mathrm{e}^{t}}}} \\ {{{0}}} & {{{\mathrm{e}^{3 t}}}} & {{{0}}} \\ {{{0}}} & {{{3 \mathrm{e}^{2 t}}}} & {{{\mathrm{e}^{t}}}} \\ \end{array} \right]
$$
C. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{2 t}}}} & {{{1 2 t \mathrm{e}^{t} \,-\, 1 2 \mathrm{e}^{t} \,+\, t \mathrm{e}^{2 t}}}} & {{{-4 \mathrm{e}^{t}}}} \\ {{{1}}} & {{{\mathrm{e}^{2 t} \,+\, \mathrm{e}^{t}}}} & {{{3 t}}} \\ {{{0}}} & {{{-3 \mathrm{e}^{t}}}} & {{{\mathrm{e}^{t}}}} \\ \end{array} \right]
$$
D. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{t}}}} & {{{2 \mathrm{e}^{t} \,+\, \mathrm{e}^{t} \,+\, t \mathrm{e}^{t}}}} & {{{-4 \mathrm{e}^{2 t} \,+\, 2 \mathrm{e}^{t}}}} \\ {{{0}}} & {{{\mathrm{e}^{3 t}}}} & {{{0}}} \\ {{{0}}} & {{{3 \mathrm{e}^{t} \,-\, 2 \mathrm{e}^{2 t}}}} & {{{\mathrm{e}^{2 t}}}} \\ \end{array} \right]
$$
E. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{2 t}}}} & {{{1 2 \mathrm{e}^{2 t} \,-\, 3 t \,+\, 1 3 t \mathrm{e}^{t}}}} & {{{-4 \,+\, 4 \mathrm{e}^{t}}}} \\ {{{0}}} & {{{\mathrm{e}^{t}}}} & {{{3 t \mathrm{e}^{t}}}} \\ {{{0}}} & {{{-2 \mathrm{e}^{t} \,+\, 3 \mathrm{e}^{2 t}}}} & {{{\mathrm{e}^{3 t}}}} \\ \end{array} \right]
$$
F. $$
\left[ \begin{array} {c c c} {{{t}}} & {{{1 2 t \mathrm{e}^{t}}}} & {{{-4 \,+\, 4 \mathrm{e}^{t}}}} \\ {{{0}}} & {{{\mathrm{e}^{t} \,+\, t}}} & {{{0}}} \\ {{{0}}} & {{{-3 \mathrm{e}^{t}}}} & {{{\mathrm{e}^{t}}}} \\ \end{array} \right]
$$
G. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{2 t}}}} & {{{1 2 \mathrm{e}^{t} \,-\, 1 2 \mathrm{e}^{2 t} \,+\, 1 3 t \mathrm{e}^{2 t}}}} & {{{-4 \mathrm{e}^{t} \,+\, 4 \mathrm{e}^{2 t}}}} \\ {{{0}}} & {{{\mathrm{e}^{2 t}}}} & {{{0}}} \\ {{{0}}} & {{{-3 \mathrm{e}^{t} \,+\, 3 \mathrm{e}^{2 t}}}} & {{{\mathrm{e}^{t}}}} \\ \end{array} \right]
$$
H. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{2 t}}}} & {{{1 2 \mathrm{e}^{t}}}} & {{{-4 \mathrm{e}^{t}}}} \\ {{{0}}} & {{{\mathrm{t}}}} & {{{3}}} \\ {{{1}}} & {{{3 t \mathrm{e}^{2 t}}}} & {{{\mathrm{e}^{t}}}} \\ \end{array} \right]
$$
I. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{t}}}} & {{{1 \,-\, 2 \mathrm{e}^{t} \,+\, t \mathrm{e}^{2 t}}}} & {{{-4 \mathrm{e}^{t}}}} \\ {{{0}}} & {{{\mathrm{e}^{2 t}}}} & {{{t \mathrm{e}^{t}}}} \\ {{{1}}} & {{{3 \mathrm{e}^{t} \,-\, \mathrm{e}^{t} \,+\, \mathrm{e}^{t}}}} & {{{\mathrm{e}^{t}}}} \\ \end{array} \right]
$$
J. $$
\left[ \begin{array} {c c c} {{{\mathrm{e}^{t}}}} & {{{1 3 \mathrm{e}^{2 t} \,-\, t \mathrm{e}^{t} \,+\, \mathrm{e}^{t}}}} & {{{-4 \mathrm{e}^{t} \,+\, 2}}} \\ {{{0}}} & {{{\mathrm{e}^{t}}}} & {{{t \mathrm{e}^{2 t}}}} \\ {{{0}}} & {{{-3 \mathrm{e}^{t} \,+\, 2 \mathrm{e}^{2 t}}}} & {{{\mathrm{e}^{2 t}}}} \\ \end{array} \right]
$$
Answer: G