## Precalculus (6th Edition) Blitzer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{7}{2} & -2 & -2 \\ -\frac{5}{2} & 1 & 2 \\ -1 & 0 & 1 \\ \end{matrix} \right]$.
Consider the given matrix $A=\left[ \begin{matrix} 2 & 4 & -4 \\ 1 & 3 & -4 \\ 2 & 4 & -3 \\ \end{matrix} \right]$. Compute matrix in the form of: $\left[ \left. A \right|I \right]$ The augment with identity matrix: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 2 & 4 & -4 & 1 & 0 & 0 \\ 1 & 3 & -4 & 0 & 1 & 0 \\ 2 & 4 & -3 & 0 & 0 & 1 \\ \end{matrix} \right]$ 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}{2}\times {{R}_{1}}, \\ & {{R}_{3}}\to {{R}_{3}}-1\times {{R}_{1}} \\ \end{align} The resulting matrix is: $\left[ \begin{matrix} 2 & 4 & -4 & 1 & 0 & 0 \\ 0 & 1 & -2 & -\frac{1}{2} & 1 & 0 \\ 0 & 0 & 1 & -1 & 0 & 1 \\ \end{matrix} \right]$ Again, apply the row operations as below: \begin{align} & {{R}_{2}}\to {{R}_{2}}+2\times {{R}_{3}}, \\ & {{R}_{1}}\to {{R}_{1}}+4\times {{R}_{3}}, \\ & {{R}_{1}}\to {{R}_{1}}-4\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 & \frac{7}{2} & -2 & -2 \\ 0 & 1 & 0 & -\frac{5}{2} & 1 & 2 \\ 0 & 0 & 1 & -1 & 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Where ${{A}^{-1}}=\left[ B \right]$ Therefore, the inverse of the matrix is; ${{A}^{-1}}=\left[ \begin{matrix} \frac{7}{2} & -2 & -2 \\ -\frac{5}{2} & 1 & 2 \\ -1 & 0 & 1 \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And, ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 2 & 4 & -4 \\ 1 & 3 & -4 \\ 2 & 4 & -3 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 2 & 4 & -4 \\ 1 & 3 & -4 \\ 2 & 4 & -3 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{7}{2} & -2 & -2 \\ -\frac{5}{2} & 1 & 2 \\ -1 & 0 & 1 \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 2\times \frac{7}{2}+4\times \left( -\frac{5}{2} \right)+\left( -4 \right)\times \left( -1 \right) & \left( -4 \right)+4+0 & \left( -4 \right)+8+\left( -4 \right) \\ 1\times \frac{7}{2}+3\times \left( -\frac{5}{2} \right)+\left( -4 \right)\times \left( -1 \right) & \left( -2 \right)+3+0 & \left( -2 \right)+6+\left( -4 \right) \\ 2\times \frac{7}{2}+4\times \left( -\frac{5}{2} \right)+\left( -3 \right)\times \left( -1 \right) & \left( -4 \right)+4+0 & \left( -4 \right)+8+\left( -3 \right) \\ \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} \frac{7}{2} & -2 & -2 \\ -\frac{5}{2} & 1 & 2 \\ -1 & 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 4 & -4 \\ 1 & 3 & -4 \\ 2 & 4 & -3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{7}{2}\times 2+\left( -2 \right)+\left( -4 \right) & \frac{7}{2}\times 4+\left( -6 \right)+\left( -8 \right) & \frac{7}{2}\times \left( -4 \right)+\left( -2 \right)\times \left( -4 \right)+\left( -2 \right)\times \left( -3 \right) \\ \left( -\frac{5}{2} \right)\times 2+1+4 & \left( -\frac{5}{2} \right)\times 4+3+8 & \left( -\frac{5}{2} \right)\times \left( -4 \right)+\left( -4 \right)+\left( -6 \right) \\ \left( -6 \right)+0+2 & \left( -4 \right)+0+4 & \left( -1 \right)\times \left( -4 \right)+0\times +\left( -3 \right) \\ \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 ${{A}^{-1}}A={{I}_{_{3}}}$.