## 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 provided expression is $A=\left[ \begin{matrix} -2 & 1 & -1 \\ -5 & 2 & -1 \\ 3 & -1 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \\ \end{matrix} \right]$ Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} -2 & 1 & -1 \\ -5 & 2 & -1 \\ 3 & -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -2 \right)\times 1+1\times 2+\left( -1 \right)\times \left( -1 \right) & \left( -2 \right)\times 0+1\times 1+\left( -1 \right)\times 1 & \left( -2 \right)\times 1+1\times 3+\left( -1 \right)\times 1 \\ \left( -5 \right)\times 1+2\times 2+\left( -1 \right)\times \left( -1 \right) & \left( -5 \right)\times 0+2\times 1+\left( -1 \right)\times 1 & \left( -5 \right)\times 1+2\times 3+\left( -1 \right)\times 1 \\ 3\times 1+\left( -1 \right)\times 2+1\times \left( -1 \right) & 3\times 0+\left( -1 \right)\times 1+1\times 1 & 3\times 1+\left( -1 \right)\times 3+1\times 1 \\ \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} 1 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -2 & 1 & -1 \\ -5 & 2 & -1 \\ 3 & -1 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times \left( -2 \right)+0\times \left( -5 \right)+1\times 3 & 1\times 1+0\times 2+1\times \left( -1 \right) & 1\times \left( -1 \right)+0\times \left( -1 \right)+1\times 1 \\ 2\times \left( -2 \right)+1\times \left( -5 \right)+3\times 3 & 2\times 1+1\times 2+3\times \left( -1 \right) & 2\times \left( -1 \right)+1\times \left( -1 \right)+3\times 1 \\ \left( -1 \right)\times \left( -2 \right)+1\times \left( -5 \right)+1\times 3 & \left( -1 \right)\times 1+1\times 2+1\times \left( -1 \right) & \left( -1 \right)\times \left( -1 \right)+1\times \left( -1 \right)+1\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 $B={{A}^{-1}}$. $B=\left[ \begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \\ \end{matrix} \right]$ And $A=\left[ \begin{matrix} -2 & 1 & -1 \\ -5 & 2 & -1 \\ 3 & -1 & 1 \\ \end{matrix} \right]$ Determine the inverse of matrix A. Now, reduce matrix to row form: $A={{\left[ \begin{matrix} -2 & 1 & -1 \\ -5 & 2 & -1 \\ 3 & -1 & 1 \\ \end{matrix} \right]}^{-1}}$ Augment with an identity matrix $A=\left[ \begin{matrix} -2 & 1 & -1 & 1 & 0 & 0 \\ -5 & 2 & -1 & 0 & 1 & 0 \\ 3 & -1 & 1 & 0 & 0 & 1 \\ \end{matrix} \right]$ Swap matrix rows: ${{R}_{1}}\leftrightarrow {{R}_{2}}$ $A=\left[ \begin{matrix} -5 & 2 & -1 & 0 & 1 & 0 \\ -2 & 1 & -1 & 1 & 0 & 0 \\ 3 & -1 & 1 & 0 & 0 & 1 \\ \end{matrix} \right]$ Again swap matrix row \begin{align} & {{R}_{2}}\leftrightarrow {{R}_{3}} \\ & {{R}_{_{2}}}\to {{R}_{2}}-\frac{2}{5}\times {{R}_{1}} \\ \end{align} And \begin{align} & {{R}_{3}}\to {{R}_{3}}+\frac{3}{5}\times {{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-1\times {{R}_{2}} \\ \end{align} $\left[ \begin{matrix} -5 & 2 & -1 & 0 & 1 & 0 \\ 0 & \frac{1}{5} & -\frac{3}{5} & 1 & -\frac{2}{5} & 0 \\ 0 & 0 & 1 & -1 & 1 & 1 \\ \end{matrix} \right]$ \begin{align} & {{R}_{2}}\to {{R}_{2}}+\frac{3}{5}\times {{R}_{3}} \\ & {{R}_{1}}\to {{R}_{1}}+1\times {{R}_{3}} \\ \end{align} $\left[ \begin{matrix} -5 & 2 & 0 & -1 & 2 & 1 \\ 0 & \frac{1}{5} & 0 & \frac{2}{5} & \frac{1}{5} & \frac{3}{5} \\ 0 & 0 & 1 & -1 & 1 & 1 \\ \end{matrix} \right]$ \begin{align} & {{R}_{1}}\to 5\times {{R}_{2}} \\ & {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{2}} \\ & {{R}_{1}}\to -\frac{1}{5}\times {{R}_{1}} \end{align} $\left[ \begin{matrix} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 2 & 1 & 3 \\ 0 & 0 & 1 & -1 & 1 & 1 \\ \end{matrix} \right]$ ${{A}^{-1}}=\left[ \begin{matrix} 1 & {} & 1 \\ 2 & 1 & 3 \\ -1 & 1 & 1 \\ \end{matrix} \right]$ Thus, $B={{A}^{-1}}$