## Precalculus (6th Edition) Blitzer

The inverse of matrix $A$ is, ${{A}^{-1}}=\left[ \begin{matrix} 8 & -8 & 5 \\ -3 & 2 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right],A{{A}^{-1}}={{I}_{3}},{{A}^{-1}}A={{I}_{3}}$
First write the augmented matrix $\left[ A|{{I}_{3}} \right]$ as follows: $A=\left[ \begin{matrix} 1 & 3 & -2 & 1 & 0 & 0 \\ 4 & 13 & -7 & 0 & 1 & 0 \\ 5 & 16 & -8 & 0 & 0 & 1 \\ \end{matrix} \right]$ 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, ${{R}_{2}}\to {{R}_{2}}-4{{R}_{1}}\text{ and }{{R}_{3}}\to {{R}_{3}}-5{{R}_{1}}$ to get, $\left[ \begin{matrix} 1 & 3 & -2 & 1 & 0 & 0 \\ 4 & 13 & -7 & 0 & 1 & 0 \\ 5 & 16 & -8 & 0 & 0 & 1 \\ \end{matrix} \right]\sim \left[ \begin{matrix} 1 & 3 & -2 & 1 & 0 & 0 \\ 0 & 1 & 1 & -4 & 1 & 0 \\ 0 & 1 & 2 & -5 & 0 & 1 \\ \end{matrix} \right]$ Apply, ${{R}_{3}}\to {{R}_{3}}-{{R}_{2}}$ to get, $\left[ \begin{matrix} 1 & 3 & -2 & 1 & 0 & 0 \\ 0 & 1 & 1 & -4 & 1 & 0 \\ 0 & 1 & 2 & -5 & 0 & 1 \\ \end{matrix} \right]\sim \left[ \begin{matrix} 1 & 3 & -2 & 1 & 0 & 0 \\ 0 & 1 & 1 & -4 & 1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 1 \\ \end{matrix} \right]$ Now apply the row transformation, ${{R}_{1}}\to {{R}_{1}}-3{{R}_{3}}\text{ and }{{R}_{2}}\to {{R}_{2}}+{{R}_{3}}$ to get, $\left[ \begin{matrix} 1 & 3 & -2 & 1 & 0 & 0 \\ 0 & 1 & 1 & -4 & 1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 1 \\ \end{matrix} \right]\sim \left[ \begin{matrix} 1 & 0 & -5 & 13 & -3 & 0 \\ 0 & 1 & 0 & -3 & 2 & -1 \\ 0 & 0 & 1 & -1 & -1 & 1 \\ \end{matrix} \right]$ Apply, ${{R}_{1}}\to {{R}_{1}}+5{{R}_{3}}$, to get, $\left[ \begin{matrix} 1 & 0 & -5 & 13 & -3 & 0 \\ 0 & 1 & 0 & -3 & 2 & -1 \\ 0 & 0 & 1 & -1 & -1 & 1 \\ \end{matrix} \right]\sim \left[ \begin{matrix} 1 & 0 & 0 & 8 & -8 & 5 \\ 0 & 1 & 0 & -3 & 2 & -1 \\ 0 & 0 & 1 & -1 & -1 & 1 \\ \end{matrix} \right]$ This is of the form $\left[ {{I}_{3}}|B \right]$ ; Where, $B=\left[ \begin{matrix} 8 & -8 & 5 \\ -3 & 2 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right]$ Therefore, $B$ is the multiplicative inverse of the matrix $A$. Thus, ${{A}^{-1}}=B=\left[ \begin{matrix} 8 & -8 & 5 \\ -3 & 2 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right]$ 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 & 3 & -2 \\ 4 & 13 & -7 \\ 5 & 16 & -8 \\ \end{matrix} \right]\left[ \begin{matrix} 8 & -8 & 5 \\ -3 & 2 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( 8 \right)+3\left( -3 \right)+\left( -2 \right)\left( -1 \right) & 1\left( -8 \right)+3\left( 2 \right)+\left( -2 \right)\left( -1 \right) & 1\left( 5 \right)+3\left( -1 \right)+\left( -2 \right)\left( 1 \right) \\ 4\left( 8 \right)+13\left( -3 \right)+\left( -7 \right)\left( -1 \right) & 4\left( -8 \right)+13\left( 2 \right)+\left( -7 \right)\left( -1 \right) & 4\left( 5 \right)+13\left( -1 \right)+\left( -7 \right)\left( 1 \right) \\ 5\left( 8 \right)+16\left( -3 \right)+\left( -8 \right)\left( -1 \right) & 5\left( -8 \right)+16\left( 2 \right)+\left( -8 \right)\left( -1 \right) & 5\left( 5 \right)+16\left( -1 \right)+\left( -8 \right)\left( 1 \right) \\ \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} 8 & -8 & 5 \\ -3 & 2 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 3 & -2 \\ 4 & 13 & -7 \\ 5 & 16 & -8 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 8\left( 1 \right)+\left( -8 \right)\left( 4 \right)+\left( 5 \right)\left( 5 \right) & 8\left( 3 \right)+\left( -8 \right)\left( 13 \right)+\left( 5 \right)\left( 16 \right) & 8\left( -2 \right)+\left( -8 \right)\left( -7 \right)+5\left( -8 \right) \\ \left( -3 \right)\left( 1 \right)+2\left( 4 \right)+\left( -1 \right)\left( 5 \right) & \left( -3 \right)\left( 3 \right)+2\left( 13 \right)+\left( -1 \right)\left( 16 \right) & \left( -3 \right)\left( -2 \right)+2\left( -7 \right)+\left( -1 \right)\left( -8 \right) \\ \left( -1 \right)\left( 1 \right)+\left( -1 \right)\left( 4 \right)+\left( 1 \right)\left( 5 \right) & \left( -1 \right)\left( 3 \right)+\left( -1 \right)\left( 13 \right)+1\left( 16 \right) & \left( -1 \right)\left( -2 \right)+\left( -7 \right)\left( 4 \right)+1\left( -8 \right) \\ \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, ${{A}^{-1}}=\left[ \begin{matrix} 8 & -8 & 5 \\ -3 & 2 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right]$.