## Precalculus (6th Edition) Blitzer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right]$.
Consider the provided matrix $A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]$. Compute matrix in the form of: $\left[ \left. A \right|I \right]$ Augment matrix with identity matrix is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 3 & 0 & 0 & 1 & 0 & 0 \\ 0 & 6 & 0 & 0 & 1 & 0 \\ 0 & 0 & 9 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, we will use row operations to reduce in row echelon form for the inverse: \begin{align} & {{R}_{3}}\to \frac{1}{9}\times {{R}_{3}}, \\ & {{R}_{2}}\to \frac{1}{6}\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 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{6} & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{9} \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Therefore, the inverse of the matrix is: ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right]$ Where $B={{A}^{-1}}$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \\ \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{1}{3} & 0 & 0 \\ 0 & \frac{1}{6} & 0 \\ 0 & 0 & \frac{1}{9} \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \\ \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 ${{A}^{-1}}A={{I}_{_{3}}}$.