## Precalculus (6th Edition) Blitzer

The inverse is $\left[ \begin{matrix} \frac{{{e}^{-x}}}{2} & -\frac{{{e}^{-3x}}}{2} \\ \frac{{{e}^{-3x}}}{2} & \frac{{{e}^{-5x}}}{2} \\ \end{matrix} \right]$.
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}