## Precalculus (6th Edition) Blitzer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]$.
Consider the given matrix $A=\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]$ Compute matrix in the form of: $\left[ \left. A \right|I \right]$ The augment with identity matrix is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & -1 & 1 & 1 & 0 & 0 \\ 0 & 2 & -1 & 0 & 1 & 0 \\ 2 & 3 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below: \begin{align} & {{R}_{1}}\leftrightarrow {{R}_{3}}, \\ & {{R}_{3}}\to {{R}_{3}}-\frac{1}{2}\times {{R}_{1}}, \\ & {{R}_{2}}\leftrightarrow {{R}_{3}}, \\ & {{R}_{3}}\to {{R}_{3}}+\frac{4}{5}\times {{R}_{2}} \\ \end{align} The resulting matrix is: $\left[ \begin{matrix} 2 & 3 & 0 & 0 & 0 & 1 \\ 0 & -\frac{5}{2} & 1 & 1 & 0 & -\frac{1}{2} \\ 0 & 0 & -\frac{1}{5} & \frac{4}{5} & 1 & -\frac{2}{5} \\ \end{matrix} \right]$ Again, apply the row operations as below: \begin{align} & {{R}_{3}}\to -5\times {{R}_{3}}, \\ & {{R}_{2}}\to {{R}_{2}}-1\times {{R}_{3}}, \\ & {{R}_{2}}\to -\frac{2}{5}\times {{R}_{2}}, \\ & {{R}_{1}}\to {{R}_{1}}-3\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 & 3 & 3 & -1 \\ 0 & 1 & 0 & -2 & -2 & 1 \\ 0 & 0 & 1 & -4 & -5 & 2 \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Where ${{A}^{-1}}=\left[ B \right]$ Therefore, the inverse of the matrix is: ${{A}^{-1}}=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 3+4+\left( -4 \right) & 3+2+\left( -5 \right) & \left( -1 \right)+\left( -1 \right)+2 \\ 0+\left( -4 \right)+4 & 0+\left( -4 \right)+5 & 0+1+\left( -2 \right) \\ 6+\left( -6 \right)+0 & 6+\left( -6 \right)+0 & \left( -2 \right)+3+0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} And, \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3+0+\left( -2 \right) & 1+0+\left( -2 \right) & 3+\left( -3 \right)+0 \\ \left( -2 \right)+0+2 & \left( -1 \right)+0+2 & \left( -2 \right)+2+0 \\ \left( -4 \right)+0+4 & \left( -4 \right)+0+4 & \left( -4 \right)+5+0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} Thus, $A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.