## Precalculus (6th Edition) Blitzer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{6} \\ \end{matrix} \right]$
Consider the given matrix $A=\left[ \begin{matrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]$ Augment with identity matrix: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 2 & 0 & 0 & 1 & 0 & 0 \\ 0 & 4 & 0 & 0 & 1 & 0 \\ 0 & 0 & 6 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, we will use following row operations to row reduce for an inverse: \begin{align} & {{R}_{3}}\to \frac{1}{6}\times {{R}_{3}}, \\ & {{R}_{2}}\to \frac{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{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{4} & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{6} \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} So, the inverse of the matrix is: ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{6} \\ \end{matrix} \right]$ Where $B={{A}^{-1}}$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]$ Now, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{6} \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 2\times \frac{1}{2}+0\times 0+0\times 0 & 2\times 0+0\times \frac{1}{4}+0\times 0 & 2\times 0+0\times 0+0\times \frac{1}{6} \\ 0\times \frac{1}{2}+4\times 0+0\times 0 & 0\times 0+4\times \frac{1}{4}+0\times 0 & 0\times 0+4\times 0+0\times \frac{1}{6} \\ 0\times \frac{1}{2}+0\times 0+6\times 0 & 0\times 0+0\times \frac{1}{4}+6\times 0 & 0\times 0+0\times 0+6\times \frac{1}{6} \\ \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} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{6} \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{1}{2}\times 2+0\times 0+0\times 0 & \frac{1}{2}\times 0+0\times 4+0\times 0 & \frac{1}{2}\times 0+0\times 0+0\times 6 \\ 0\times 2+\frac{1}{4}\times 0+0\times 0 & 0\times 0+\frac{1}{4}\times 4+0\times 0 & 0\times 0+\frac{1}{4}\times 0+0\times 6 \\ 0\times 2+0\times 0+\frac{1}{6}\times 0 & 0\times 0+0\times 4+\frac{1}{6}\times 0 & 0\times 0+0\times 0+\frac{1}{6}\times 6 \\ \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 ${{A}^{-1}}A={{I}_{_{3}}}$