## Precalculus (6th Edition) Blitzer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 3 & 2 & 6 \\ \end{matrix} \right]$.
Consider the given matrix $A=\left[ \begin{matrix} 2 & 2 & -1 \\ 0 & 3 & -1 \\ -1 & -2 & 1 \\ \end{matrix} \right]$. Compute matrix in the form of: $\left[ \left. A \right|I \right]$ The augment matrix with identity matrix is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 2 & 2 & -1 & 1 & 0 & 0 \\ 0 & 3 & -1 & 0 & 1 & 0 \\ -1 & -2 & 1 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below: \begin{align} & {{R}_{3}}\to {{R}_{3}}+\frac{1}{2}\times {{R}_{1}}, \\ & {{R}_{3}}\to {{R}_{3}}+\frac{1}{3}\times {{R}_{2}} \\ \end{align} The resulting matrix is: $\left[ \begin{matrix} 2 & 2 & -1 & 1 & 0 & 0 \\ 0 & 3 & -1 & 0 & 1 & 0 \\ 0 & 0 & \frac{1}{6} & \frac{1}{2} & \frac{1}{3} & 1 \\ \end{matrix} \right]$ Again, apply the row operations as below: \begin{align} & {{R}_{3}}\to 6\times {{R}_{3}}, \\ & {{R}_{2}}\to {{R}_{2}}+1\times {{R}_{3}}, \\ & {{R}_{1}}\to {{R}_{1}}+1\times {{R}_{2}}, \\ & {{R}_{2}}\to \frac{1}{3}\times {{R}_{2}}, \\ & {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{2}}, \\ & {{R}_{1}}\to \frac{1}{2}\times {{R}_{1}} \\ \end{align} The resulting matrix is: \begin{align} & \left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 3 & 2 & 6 \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Where ${{A}^{-1}}=\left[ B \right]$ Therefore, the inverse of the matrix is: ${{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 3 & 2 & 6 \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 2 & 2 & -1 \\ 0 & 3 & -1 \\ -1 & -2 & 1 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 2 & 2 & -1 \\ 0 & 3 & -1 \\ -1 & -2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 3 & 2 & 6 \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 2+2+\left( -3 \right) & 0+2+\left( -2 \right) & 2+4+\left( -6 \right) \\ 0+3+\left( -3 \right) & 0+3+\left( -2 \right) & 0+6+\left( -6 \right) \\ \left( -1 \right)+\left( -2 \right)+3 & 0+\left( -2 \right)+2 & \left( -1 \right)+\left( -4 \right)+6 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ \end{matrix} \right] \\ & A{{A}^{-1}}={{I}_{3}} \end{align} And, \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 3 & 2 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 2 & -1 \\ 0 & 3 & -1 \\ -1 & -2 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2+0+1 & 2+0+\left( -2 \right) & \left( -1 \right)+0+1 \\ 2+0+\left( -2 \right) & 2+3+\left( -4 \right) & \left( -1 \right)+\left( -1 \right)+2 \\ 6+0+\left( -6 \right) & 6+6+\left( -12 \right) & \left( -3 \right)+\left( -2 \right)+6 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & {{A}^{-1}}A={{I}_{3}} \end{align} Thus, $A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.