Precalculus (6th Edition) Blitzer

The inverse is, ${{C}^{-1}}=\frac{1}{5}\left[ \begin{array}{*{35}{r}} 3 & -2 \\ 1 & 1 \\ \end{array} \right]$
Consider the inverse of the matrix $C$ ${{C}^{-1}}=\frac{1}{\left| C \right|}\text{adj}\left( C \right)$ Consider the determinant of the matrix \begin{align} & \left| C \right|=\left| \begin{array}{*{35}{r}} 1 & 2 \\ -1 & 3 \\ \end{array} \right| \\ & =3+2 \\ & =5 \end{align} Consider the adjoin of the matrix \begin{align} & \text{adj}\left( C \right)={{\left[ \begin{matrix} +\left( 3 \right) & -\left( -1 \right) \\ -\left( 2 \right) & +\left( 1 \right) \\ \end{matrix} \right]}^{t}} \\ & ={{\left[ \begin{array}{*{35}{r}} 3 & 1 \\ -2 & 1 \\ \end{array} \right]}^{t}} \\ & =\left[ \begin{array}{*{35}{r}} 3 & -2 \\ 1 & 1 \\ \end{array} \right] \end{align} Therefore, the inverse of the matrix $C$ is given by, ${{C}^{-1}}=\frac{1}{5}\left[ \begin{array}{*{35}{r}} 3 & -2 \\ 1 & 1 \\ \end{array} \right]$ Therefore, the inverse of the matrix is ${{C}^{-1}}=\frac{1}{5}\left[ \begin{array}{*{35}{r}} 3 & -2 \\ 1 & 1 \\ \end{array} \right]$.