Answer
$\begin{array}( \\ \frac{1}{2e^{4x}} \\ \\ \end{array}
\begin{bmatrix} e^{3x} & e^{2x} \\-e^{2x} & e^{x} \end{bmatrix} $
for all real $x$.
Work Step by Step
Step 1. Given the matrix:
$\begin{array}( \\A= \\ \\ \end{array}
\begin{bmatrix} e^x & -e^{2x} \\e^{2x} & e^{3x} \end{bmatrix} $
Step 2. Use the formula on page 725 to get:
$\begin{array}( \\A^{-1}=\frac{1}{e^xe^{3x}+e^{2x}e^{2x}} \\ \\ \end{array}
\begin{bmatrix} e^{3x} & e^{2x} \\-e^{2x} & e^{x} \end{bmatrix}\begin{array}( \\ =\frac{1}{2e^{4x}} \\ \\ \end{array}
\begin{bmatrix} e^{3x} & e^{2x} \\-e^{2x} & e^{x} \end{bmatrix} $
Apparently, this inverse is valid for all real $x$ values.
