Answer
The inverse is $\left[ \begin{matrix}
\frac{{{e}^{-x}}}{2} & -\frac{{{e}^{-3x}}}{2} \\
\frac{{{e}^{-3x}}}{2} & \frac{{{e}^{-5x}}}{2} \\
\end{matrix} \right]$.
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,
$\begin{align}
& a={{e}^{x}} \\
& b={{e}^{3x}} \\
& c=-{{e}^{3x}} \\
& d={{e}^{5x}}
\end{align}$
Now, the inverse is given by, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right]$
Substitute the values to get, $\begin{align}
& {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right]\Bbbk \\
& =\frac{1}{2{{e}^{6x}}}\left[ \begin{matrix}
{{e}^{5x}} & -{{e}^{3x}} \\
{{e}^{3x}} & {{e}^{x}} \\
\end{matrix} \right] \\
& =\frac{1}{2}\left[ \begin{matrix}
\frac{{{e}^{5x}}}{{{e}^{6x}}} & -\frac{{{e}^{3x}}}{{{e}^{6x}}} \\
\frac{{{e}^{3x}}}{{{e}^{6x}}} & \frac{{{e}^{x}}}{{{e}^{6x}}} \\
\end{matrix} \right] \\
& =\frac{1}{2}\left[ \begin{matrix}
{{e}^{-x}} & -{{e}^{-3x}} \\
{{e}^{-3x}} & {{e}^{-5x}} \\
\end{matrix} \right]
\end{align}$
Therefore, the value of the inverse of the matrix is given by,
${{A}^{-1}}=\left[ \begin{matrix}
\frac{{{e}^{-x}}}{2} & -\frac{{{e}^{-3x}}}{2} \\
\frac{{{e}^{-3x}}}{2} & \frac{{{e}^{-5x}}}{2} \\
\end{matrix} \right]$
Therefore, the inverse of the matrix $ A $ is $\left[ \begin{matrix}
\frac{{{e}^{-x}}}{2} & -\frac{{{e}^{-3x}}}{2} \\
\frac{{{e}^{-3x}}}{2} & \frac{{{e}^{-5x}}}{2} \\
\end{matrix} \right]$
Now, check the result, to show that $ A{{A}^{-1}}=I $, $\begin{align}
& \left[ \begin{matrix}
{{e}^{x}} & {{e}^{3x}} \\
-{{e}^{3x}} & {{e}^{5x}} \\
\end{matrix} \right]\left[ \begin{matrix}
\frac{{{e}^{-x}}}{2} & -\frac{{{e}^{-3x}}}{2} \\
\frac{{{e}^{-3x}}}{2} & \frac{{{e}^{-5x}}}{2} \\
\end{matrix} \right]=\left[ \begin{matrix}
{{e}^{x}}\frac{{{e}^{-x}}}{2}+{{e}^{3x}}\frac{{{e}^{-3x}}}{2} & {{e}^{x}}-\frac{{{e}^{3x}}}{2}+{{e}^{3x}}\frac{{{e}^{-5x}}}{2} \\
-{{e}^{3x}}\frac{{{e}^{-x}}}{2}+{{e}^{5x}}\frac{{{e}^{-3x}}}{2} & -{{e}^{3x}}-\frac{{{e}^{-3x}}}{2}+{{e}^{5x}}\frac{{{e}^{-5x}}}{2} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]
\end{align}$
