## Precalculus (6th Edition) Blitzer

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