## Precalculus (6th Edition) Blitzer

Consider the above statement such that the Gauss-Jordan elimination method can be used to find the multiplicative inverse of the $3\times 3$ matrix if the determinant of the given matrix is nonzero. Example: Consider the following matrix: $A=\left[ \begin{matrix} 5 & 0 & 2 \\ 2 & 2 & 1 \\ -3 & 1 & -1 \\ \end{matrix} \right]$ Calculate the determinant of the above matrix as below: \begin{align} & \left| A \right|=5\left( -2-1 \right)+0\left( -2-3 \right)+2\left( 2+6 \right) \\ & =-15+16 \\ & =1 \\ & \ne 0 \end{align} Therefore, the inverse of the matrix exists. Apply Gauss--Jordan elimination method: The augmented matrix $\left[ A|{{I}_{3}} \right]$ is $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 0 & 2 & 1 & 0 & 0 \\ 2 & 2 & 1 & 0 & 1 & 0 \\ -3 & 1 & -1 & 0 & 0 & 1 \\ \end{matrix} \right]$ Use the elementary row operations on $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]$ to get a matrix of the form $\left[ \begin{matrix} {{I}_{3}} & B \\ \end{matrix} \right]$. The matrix B is inverse of A, that is, ${{A}^{-1}}=B$. ${{R}_{1}}\to \frac{1}{5}{{R}_{1}}$ gives $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\ 2 & 2 & 1 & 0 & 1 & 0 \\ -3 & 1 & -1 & 0 & 0 & 1 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\left( -2 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( 3 \right){{R}_{1}}$ gives $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 2 & \frac{1}{5} & -\frac{2}{5} & 1 & 0 \\ 0 & 1 & -\frac{1}{5} & \frac{3}{5} & 0 & 1 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{2}{{R}_{2}}$ gives $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & \frac{1}{10} & -\frac{1}{5} & \frac{1}{2} & 0 \\ 0 & 1 & -\frac{1}{5} & \frac{3}{5} & 0 & 1 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\left( -1 \right){{R}_{2}}$ gives $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & \frac{1}{10} & -\frac{1}{5} & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{10} & \frac{4}{5} & -\frac{1}{2} & 1 \\ \end{matrix} \right]$ ${{R}_{3}}\to 10{{R}_{3}}$ gives $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & \frac{2}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & \frac{1}{10} & -\frac{1}{5} & \frac{1}{2} & 0 \\ 0 & 0 & 1 & 8 & -5 & 10 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}+\left( -\frac{2}{5} \right){{R}_{3}},{{R}_{2}}\to {{R}_{2}}+\left( -\frac{1}{10} \right){{R}_{3}}$ gives $\left[ \begin{matrix} A & {{I}_{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 & -3 & 2 & -4 \\ 0 & 1 & 0 & -1 & 1 & -1 \\ 0 & 0 & 1 & 8 & -5 & 10 \\ \end{matrix} \right]$ That is $\left[ \begin{matrix} {{I}_{3}} & B \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 & -3 & 2 & -4 \\ 0 & 1 & 0 & -1 & 1 & -1 \\ 0 & 0 & 1 & 8 & -5 & 10 \\ \end{matrix} \right]$ That is ${{A}^{-1}}=B=\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right]$ \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 5 & 0 & 2 \\ 2 & 2 & 1 \\ -3 & 1 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right]\left[ \begin{matrix} 5 & 0 & 2 \\ 2 & 2 & 1 \\ -3 & 1 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} Hence ${{A}^{-1}}=\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right],A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{3}}$