## Precalculus (6th Edition) Blitzer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \\ \end{matrix} \right]$.
Consider the given matrix $A=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \\ \end{matrix} \right]$. Compute matrix in the form of: $\left[ \left. A \right|I \right]$ The augment matrix with identity matrix is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 3 & 2 & 6 & 1 & 0 & 0 \\ 1 & 1 & 2 & 0 & 1 & 0 \\ 2 & 2 & 5 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below: \begin{align} & {{R}_{2}}\to {{R}_{2}}-\frac{1}{3}\times {{R}_{1}}, \\ & {{R}_{3}}\to {{R}_{3}}-\frac{2}{3}\times {{R}_{1}}, \\ & {{R}_{2}}\leftrightarrow {{R}_{_{3}}}, \\ & {{R}_{3}}\to {{R}_{3}}-\frac{1}{2}\times {{R}_{2}} \\ \end{align} The resulting matrix is: $\left[ \begin{matrix} 3 & 2 & 6 & 1 & 0 & 0 \\ 0 & \frac{2}{3} & 1 & -\frac{2}{3} & 0 & 1 \\ 0 & 0 & -\frac{1}{2} & 0 & 1 & -\frac{1}{2} \\ \end{matrix} \right]$ Again, apply the row operations as below: \begin{align} & {{R}_{3}}\to -2\times {{R}_{3}}, \\ & {{R}_{2}}\to {{R}_{2}}-1\times {{R}_{3}}, \\ & {{R}_{1}}\to {{R}_{1}}-6\times {{R}_{3}}, \\ & {{R}_{2}}\to \frac{3}{2}\times {{R}_{2}}, \\ & {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{2}}, \\ & {{R}_{1}}\to \frac{1}{3}\times {{R}_{1}} \\ \end{align} The resulting matrix is: \begin{align} & \left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & 1 & 2 & -2 \\ 0 & 1 & 0 & -1 & 3 & 0 \\ 0 & 0 & 1 & 0 & -2 & 1 \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Where ${{A}^{-1}}=\left[ B \right]$ So, The inverse of matrix is: ${{A}^{-1}}=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 3+\left( -2 \right)+0 & 6+6+\left( -12 \right) & \left( -6 \right)+0+6 \\ 1+\left( -1 \right)+0 & 2+3+\left( -4 \right) & \left( -2 \right)+0+2 \\ 2+\left( -2 \right)+0 & 4+6+\left( -10 \right) & \left( -4 \right)+0+5 \\ \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} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3+2+\left( -4 \right) & 2+2+\left( -4 \right) & 6+4+\left( -10 \right) \\ \left( -3 \right)+3+0 & \left( -2 \right)+3+0 & \left( -6 \right)+6+0 \\ 0+\left( -2 \right)+2 & 0+\left( -2 \right)+2 & 0+\left( -4 \right)+5 \\ \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 ${{A}^{-1}}A={{I}_{_{3}}}$.