## Precalculus (6th Edition) Blitzer

The inverse is, $\left[ \begin{matrix} \frac{1}{8} & \frac{5}{8} \\ \frac{3}{8} & \frac{7}{8} \\ \end{matrix} \right]$.
Consider the given matrix $A=\left[ \begin{matrix} 8 & -5 \\ -3 & 2 \\ \end{matrix} \right]$. And, identity matrix is $I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ Then, \begin{align} & \left[ I-A \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} 8 & -5 \\ -3 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1-8 & 0+5 \\ 0+3 & 1-2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -7 & 5 \\ 3 & -1 \\ \end{matrix} \right] \end{align} Now, the inverse of matrix $\left[ I-A \right]$ 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=-7 \\ & b=5 \\ & c=3 \\ & d=-1 \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}{-8}\left[ \begin{matrix} -1 & -5 \\ -3 & -7 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{1}{8} & \frac{5}{8} \\ \frac{3}{8} & \frac{7}{8} \\ \end{matrix} \right] \end{align} Therefore, the inverse of the matrix, ${{\left[ I-A \right]}^{-1}}=\left[ \begin{matrix} \frac{1}{8} & \frac{5}{8} \\ \frac{3}{8} & \frac{7}{8} \\ \end{matrix} \right]$.