## Precalculus (6th Edition) Blitzer

First, substitute the value of A in $A=IA$, and then simplify the equation such that: \begin{align} & A=IA \\ & A{{A}^{-1}}=IA{{A}^{-1}} \\ & A{{A}^{-1}}=I \end{align}
Let $A$ be the matrix, $A=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \\ \end{matrix} \right]$ Substituting, $A=IA$ we get, Where, $I=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$ Now by the using elementary transformations to find the inverse of the matrix, $A=AI$ we get, $A=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$ Replace the 1st row with the 2nd row ${{R}_{1}}\leftrightarrow {{R}_{2}}$ to get, $\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]A$ Replace the 3rd row as below, ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$ $\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 3 & 1 \\ \end{matrix} \right]A$ Further replacements: \begin{align} & {{R}_{1}}\to {{R}_{1}}-2{{R}_{2}} \\ & {{R}_{3}}\to {{R}_{3}}+5{{R}_{2}} \end{align} Simplify the matrices with the replacements in the elements to get, $\left[ \begin{matrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \\ \end{matrix} \right]A$ Also, substituting the rows for the algebraic subtraction and addition on the respective rows, we get: \begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \\ & {{R}_{1}}\to {{R}_{1}}+\frac{1}{2}{{R}_{3}} \end{align} Simplifying the matrices with the above transformations, we get: $\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ 5 & -3 & 1 \\ \end{matrix} \right]A$ The third row is divided by 2. ${{R}_{3}}\to \frac{{{R}_{3}}}{2}$ \begin{align} & \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \\ \end{matrix} \right]A \\ & I=\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \\ \end{matrix} \right]A \\ & {{A}^{-1}}I=\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \\ \end{matrix} \right]A \end{align} Thus, the invertible matrix of the $3\times 3$ multiplicative inverse matrix ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \\ \end{matrix} \right]$