## Precalculus (6th Edition) Blitzer

The inverse is $\left[ \begin{matrix} \frac{1}{2{{e}^{2x}}} & \frac{1}{2{{e}^{3x}}} \\ -\frac{1}{2{{e}^{x}}} & \frac{1}{2{{e}^{2x}}} \\ \end{matrix} \right]$.
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}