Answer
The inverse is $\left[ \begin{matrix}
\frac{1}{2{{e}^{2x}}} & \frac{1}{2{{e}^{3x}}} \\
-\frac{1}{2{{e}^{x}}} & \frac{1}{2{{e}^{2x}}} \\
\end{matrix} \right]$.
Work Step by Step
Consider the given matrix, $ A=\left[ \begin{matrix}
{{e}^{x}} & {{e}^{3x}} \\
-{{e}^{3x}} & {{e}^{5x}} \\
\end{matrix} \right]$
The inverse of matrix $ A=\left[ \begin{matrix}
a & b \\
c & d \\
\end{matrix} \right]$ is equal to ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-d & a \\
\end{matrix} \right]$
Now, we will compare the matrix to the original matrix, to get,
$\begin{align}
& a={{e}^{2x}} \\
& b={{e}^{-x}} \\
& c={{e}^{3x}} \\
& d={{e}^{2x}}
\end{align}$
Now, the inverse is, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right]$
Substitutes the values to get, $\begin{align}
& {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right] \\
& =\frac{1}{2{{e}^{4x}}}\left[ \begin{matrix}
{{e}^{2x}} & -\left( -{{e}^{x}} \right) \\
-{{e}^{3x}} & {{e}^{2x}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{{{e}^{2x}}}{2{{e}^{4x}}} & -\left( \frac{-{{e}^{x}}}{2{{e}^{4x}}} \right) \\
\frac{-{{e}^{3x}}}{2{{e}^{4x}}} & \frac{{{e}^{2x}}}{2{{e}^{4x}}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{1}{2}{{e}^{2x}} & \frac{1}{2}{{e}^{3x}} \\
-\frac{1}{2}{{e}^{x}} & \frac{1}{2}{{e}^{2x}} \\
\end{matrix} \right]
\end{align}$
Therefore, the inverse of the matrix $ A $ is $\left[ \begin{matrix}
\frac{1}{2{{e}^{2x}}} & \frac{1}{2{{e}^{3x}}} \\
-\frac{1}{2{{e}^{x}}} & \frac{1}{2{{e}^{2x}}} \\
\end{matrix} \right]$
Now, check the result: show that $ A{{A}^{-1}}=I $
$\begin{align}
& \left[ \begin{matrix}
{{e}^{2x}} & -{{e}^{x}} \\
{{e}^{3x}} & {{e}^{2x}} \\
\end{matrix} \right]\left[ \begin{matrix}
\frac{1}{2{{e}^{2x}}} & \frac{1}{2{{e}^{3x}}} \\
-\frac{1}{2{{e}^{x}}} & \frac{1}{2{{e}^{2x}}} \\
\end{matrix} \right]=\left[ \begin{matrix}
{{e}^{2x}}\frac{1}{2{{e}^{2x}}}+\left( -{{e}^{x}} \right)\left( -\frac{1}{2{{e}^{x}}} \right) & {{e}^{2x}}\frac{1}{2{{e}^{3x}}}+\left( -{{e}^{x}} \right)\frac{1}{2{{e}^{2x}}} \\
{{e}^{3x}}\frac{1}{2{{e}^{2x}}}+{{e}^{2x}}\left( -\frac{1}{2{{e}^{x}}} \right) & {{e}^{3x}}\frac{1}{2{{e}^{2x}}}+{{e}^{2x}}\frac{1}{2{{e}^{2x}}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]
\end{align}$
