Precalculus (6th Edition) Blitzer

The product of $AB={{I}_{3}}$, product of $BA={{I}_{3}}$ and $B$ is the multiplicative inverse of $A$ $B={{A}^{-1}}$. The given expression is $A=\left[ \begin{matrix} 0 & 2 & 0 \\ 3 & 3 & 2 \\ 2 & 5 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} -3.5 & -1 & 2 \\ 0.5 & 0 & 0 \\ 4.5 & 2 & -3 \\ \end{matrix} \right]$
Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} 0 & 2 & 0 \\ 3 & 3 & 2 \\ 2 & 5 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -3.5 & -1 & 2 \\ 0.5 & 0 & 0 \\ 4.5 & 2 & -3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0\times \left( -3.5 \right)+2\times 0.5+0\times 4.5 & 0\times \left( -1 \right)+2\times 0+0\times 2 & 0\times 2+2\times 0+0\times -3 \\ 3\times \left( -3.5 \right)+3\times 0.5+2\times 4.5 & 3\times \left( -1 \right)+3\times 0+2\times 2 & 3\times 2+3\times 0+2\times \left( -3 \right) \\ 2\times \left( -3.5 \right)+5\times 0.5+1\times 4.5 & 2\times \left( -1 \right)+5\times 0+1\times 2 & 2\times 2+5\times 0+1\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} Now, we will compute the matrix as $\left[ BA \right]$ \begin{align} & BA=\left[ \begin{matrix} -3.5 & -1 & 2 \\ 0.5 & 0 & 0 \\ 4.5 & 2 & -3 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 2 & 0 \\ 3 & 3 & 2 \\ 2 & 5 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -3.5 \right)\times 0+\left( -1 \right)\times 3+2\times 2 & \left( -3.5 \right)\times 2+\left( -1 \right)\times 3+2\times 5 & \left( -3.5 \right)\times 0+\left( -1 \right)\times 2+2\times 1 \\ 0.5\times 0+0\times 3+0\times 2 & 0.5\times 2+0\times 3+0\times 5 & 0.5\times 0+0\times 2+0\times 1 \\ 4.5\times 0+2\times 3+\left( -3 \right)\times 2 & 4.5\times 2+2\times 3+\left( -3 \right)\times 5 & 4.5\times 0+2\times 2+\left( -3 \right)\times 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} Both matrix $AB$ and matrix $BA$ are equal to the identity matrix Now, we will compute the matrix $B={{A}^{-1}}$. $B=\left[ \begin{matrix} -3.5 & -1 & 2 \\ 0.5 & 0 & 0 \\ 4.5 & 2 & -3 \\ \end{matrix} \right]$ And $A=\left[ \begin{matrix} 0 & 2 & 0 \\ 3 & 3 & 2 \\ 2 & 5 & 1 \\ \end{matrix} \right]$ Determine the inverse of matrix A. Now, Reduce matrix to row form: $A={{\left[ \begin{matrix} 0 & 2 & 0 \\ 3 & 3 & 2 \\ 2 & 5 & 1 \\ \end{matrix} \right]}^{-1}}$ Augment with an identity matrix, $A=\left[ \begin{matrix} 0 & 2 & 0 & 1 & 0 & 0 \\ 3 & 3 & 2 & 0 & 1 & 0 \\ 2 & 5 & 1 & 0 & 0 & 1 \\ \end{matrix} \right]$ Swap matrix rows: \begin{align} & {{R}_{1}}\leftrightarrow {{R}_{2}} \\ & {{R}_{3}}\to {{R}_{3}}-\frac{2}{3}\times {{R}_{1}} \\ & {{R}_{2}}\leftrightarrow {{R}_{3}} \\ & {{R}_{3}}\to {{R}_{3}}-\frac{2}{3}\times {{R}_{2}} \end{align} $A=\left[ \begin{matrix} 3 & 3 & 2 & 0 & 1 & 0 \\ 0 & 3 & -\frac{1}{3} & 0 & -\frac{2}{3} & 1 \\ 0 & 0 & \frac{2}{9} & 1 & \frac{4}{9} & -\frac{2}{3} \\ \end{matrix} \right]$ Again \begin{align} & {{R}_{3}}\to \frac{9}{2}\times {{R}_{3}} \\ & {{R}_{2}}\to {{R}_{2}}+\frac{1}{3}\times {{R}_{3}} \\ & {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{3}} \\ \end{align} $\left[ \begin{matrix} 3 & 3 & 0 & -9 & -3 & 6 \\ 0 & 3 & 0 & \frac{3}{2} & 0 & 0 \\ 0 & 0 & 1 & \frac{9}{2} & 2 & -3 \\ \end{matrix} \right]$ And, \begin{align} & {{R}_{2}}\to \frac{1}{3}\times {{R}_{2}} \\ & {{R}_{1}}\to {{R}_{1}}-3\times {{R}_{2}} \\ & {{R}_{1}}\to \frac{1}{3}\times {{R}_{1}} \\ \end{align} $\left[ \begin{matrix} 1 & 0 & 0 & -\frac{7}{2} & -1 & 2 \\ 0 & 1 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & \frac{9}{2} & 2 & -3 \\ \end{matrix} \right]$ Thus, ${{A}^{-1}}=\left[ \begin{matrix} -3.5 & -1 & 2 \\ 0.5 & 0 & 0 \\ 4.5 & 2 & -3 \\ \end{matrix} \right]$ Therefore, matrix $B$ is the multiplicative inverse of matrix $A$