## Precalculus (6th Edition) Blitzer

The inverse of matrix $A$ is, ${{A}^{-1}}=\left[ \begin{matrix} 3 & 0 & -2 \\ -6 & 1 & 4 \\ 1 & 0 & -1 \\ \end{matrix} \right]$ and $A{{A}^{-1}}={{I}_{3}},{{A}^{-1}}A={{I}_{3}}$
First write the augmented matrix $\left[ A|{{I}_{3}} \right]$ as below: $\left[ \begin{matrix} 1 & 0 & -2 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & -3 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now perform the elementary row operation to get a matrix of the form $\left[ {{I}_{3}}|B \right]$; then $B$ will be the inverse of the matrix $A$. So apply, ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\text{ and }{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ to get, $\left[ \begin{matrix} 1 & 0 & -2 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & -3 & 0 & 0 & 1 \\ \end{matrix} \right]\sim \left[ \begin{matrix} 1 & 0 & -2 & 1 & 0 & 0 \\ 0 & 1 & 4 & -2 & 1 & 0 \\ 0 & 0 & -1 & -1 & 0 & 1 \\ \end{matrix} \right]$ Apply, ${{R}_{3}}\to \left( -1 \right){{R}_{3}}$ to get, $\left[ \begin{matrix} 1 & 0 & -2 & 1 & 0 & 0 \\ 0 & 1 & 4 & -2 & 1 & 0 \\ 0 & 0 & -1 & -1 & 0 & 1 \\ \end{matrix} \right]\sim \left[ \begin{matrix} 1 & 0 & -2 & 1 & 0 & 0 \\ 0 & 1 & 4 & -2 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & -1 \\ \end{matrix} \right]$ Now apply the row transformation, ${{R}_{1}}\to {{R}_{1}}+2{{R}_{3}}\text{ and }{{R}_{2}}\to {{R}_{2}}-4{{R}_{3}}$ to get, $\left[ \begin{matrix} 1 & 0 & -2 & 1 & 0 & 0 \\ 0 & 1 & 4 & -2 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & -1 \\ \end{matrix} \right]\sim \left[ \begin{matrix} 1 & 0 & 0 & 3 & 0 & -2 \\ 0 & 1 & 0 & -6 & 1 & 4 \\ 0 & 0 & 1 & 1 & 0 & -1 \\ \end{matrix} \right]$ This is of the form $\left[ {{I}_{3}}|B \right]$ ; Where, $B=\left[ \begin{matrix} 3 & 0 & -2 \\ -6 & 1 & 4 \\ 1 & 0 & -1 \\ \end{matrix} \right]$ So, $B$ is the multiplicative inverse of the matrix $A$. Thus, ${{A}^{-1}}=B=\left[ \begin{matrix} 3 & 0 & -2 \\ -6 & 1 & 4 \\ 1 & 0 & -1 \\ \end{matrix} \right]$ Now verify this by showing that $A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{3}}$. So consider, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & -2 \\ 2 & 1 & 0 \\ 1 & 0 & -3 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 0 & -2 \\ -6 & 1 & 4 \\ 1 & 0 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( 3 \right)+0\left( -6 \right)+\left( -2 \right)1 & 1\left( 0 \right)+0\left( 1 \right)+\left( -2 \right)0 & 1\left( -2 \right)+0\left( 4 \right)+\left( -2 \right)\left( -1 \right) \\ 2\left( 3 \right)+1\left( -6 \right)+0\left( 1 \right) & 2\left( 0 \right)+1\left( 1 \right)+0\left( 0 \right) & 2\left( -2 \right)+1\left( 4 \right)+0\left( -1 \right) \\ 1\left( 3 \right)+0\left( -6 \right)+\left( -3 \right)\left( 1 \right) & 1\left( 0 \right)+0\left( 1 \right)+\left( -3 \right)0 & 1\left( -2 \right)+0\left( 4 \right)+\left( -3 \right)\left( -1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3-2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2+3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \end{align} Next consider, \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} 3 & 0 & -2 \\ -6 & 1 & 4 \\ 1 & 0 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & -2 \\ 2 & 1 & 0 \\ 1 & 0 & -3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3\left( 1 \right)+0\left( 2 \right)+\left( -2 \right)1 & 3\left( 0 \right)+0\left( 1 \right)+\left( -2 \right)\left( 0 \right) & 3\left( -2 \right)+0\left( 0 \right)+\left( -2 \right)\left( -3 \right) \\ \left( -6 \right)\left( 1 \right)+1\left( 2 \right)+4\left( 1 \right) & \left( -6 \right)\left( 0 \right)+1\left( 1 \right)+4\left( 0 \right) & \left( -6 \right)\left( -2 \right)+1\left( 0 \right)+4\left( -3 \right) \\ 1\left( 1 \right)+0\left( 2 \right)+\left( -1 \right)\left( 1 \right) & 1\left( 0 \right)+0\left( 1 \right)+\left( -1 \right)\left( 0 \right) & 1\left( -2 \right)+0\left( 0 \right)+\left( -3 \right)\left( -3 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3-2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2+3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \end{align} Thus, $A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{3}}$. Hence, ${{A}^{-1}}=\left[ \begin{matrix} 3 & 0 & -2 \\ -6 & 1 & 4 \\ 1 & 0 & -1 \\ \end{matrix} \right]$