## College Algebra (6th Edition)

$A^{-1}=\left[\begin{array}{ll} e^{-x}/2 & -e^{-3x}/2\\ e^{-3x}/2 & e^{-5x}/2 \end{array}\right]$
If $A=\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]$, then $A^{-1}=\displaystyle \frac{1}{ad-bc}\left[\begin{array}{ll} d & -b\\ -c & a \end{array}\right]$. ---------------------- $ad-bc=e^{x}e^{5x}-e^{3x}(-e^{3x})=e^{6x}+e^{6x}=2e^{6x}$ $A^{-1}=\displaystyle \frac{1}{2e^{6x}}\left[\begin{array}{ll} e^{5x} & -e^{3x}\\ e^{3x} & e^{x} \end{array}\right]=\displaystyle \qquad$ ... $\displaystyle \frac{e^{M}}{e^{N}}=e^{M-N}$ $=\left[\begin{array}{ll} e^{5x-6x}/2 & -e^{3x-6x}/2\\ e^{3x-6x}/2 & e^{x-6x}/2 \end{array}\right]$ $A^{-1}=\left[\begin{array}{ll} e^{-x}/2 & -e^{-3x}/2\\ e^{-3x}/2 & e^{-5x}/2 \end{array}\right]$ Check: $AA^{-1}=\left[\begin{array}{ll} e^{x} & e^{3x}\\ -e^{3x} & e^{5x} \end{array}\right]\left[\begin{array}{ll} e^{-x}/2 & -e^{-3x}/2\\ e^{-3x}/2 & e^{-5x}/2 \end{array}\right]$ $=\left[\begin{array}{ll} e^{0}/2+e^{0}/2 & -e^{-2x}/2+e^{-2x}/2\\ -e^{-2x}/2+e^{-2x}/2 & e^{0}/2+e^{0}/2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]$