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 $A=\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right],B=\left[ \begin{matrix} \frac{7}{2} & -3 & \frac{1}{2} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ \end{matrix} \right]$ Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{7}{2} & -3 & \frac{1}{2} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times \frac{7}{2}+2\times \left( -\frac{1}{2} \right)+3\times \left( -\frac{1}{2} \right) & 1\times \left( -3 \right)+2\times 0+3\times 1 & 1\times \frac{1}{2}+2\times \frac{1}{2}+3\times \left( -\frac{1}{2} \right) \\ 1\times \frac{7}{2}+3\times \left( -\frac{1}{2} \right)+4\times \left( -\frac{1}{2} \right) & 1\times \left( -3 \right)+3\times 0+4\times 1 & 1\times \frac{1}{2}+3\times \frac{1}{2}+4\times \left( -\frac{1}{2} \right) \\ 1\times \frac{7}{2}+4\times \left( -\frac{1}{2} \right)+3\times \left( -\frac{1}{2} \right) & 1\times \left( -3 \right)+4\times 0+3\times 1 & 1\times \frac{1}{2}+4\times \frac{1}{2}+3\times \left( -\frac{1}{2} \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} \frac{7}{2} & -3 & \frac{1}{2} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{7}{2}\times 1+\left( -3 \right)\times 1+\frac{1}{2}\times 1 & \frac{7}{2}\times 2+\left( -3 \right)\times 3+\frac{1}{2}\times 4 & \frac{7}{2}\times 3+\left( -3 \right)\times 4+\frac{1}{2}\times 3 \\ \left( -\frac{1}{2} \right)\times 1+0\times 1+\frac{1}{2}\times 1 & \left( -\frac{1}{2} \right)\times 2+0\times 3+\frac{1}{2}\times 4 & \left( -\frac{1}{2} \right)\times 3+0\times 4+\frac{1}{2}\times 3 \\ \left( -\frac{1}{2} \right)\times 1+1\times 1+\left( -\frac{1}{2} \right)\times 1 & \left( -\frac{1}{2} \right)\times 2+1\times 3+\left( -\frac{1}{2} \right)\times 4 & \left( -\frac{1}{2} \right)\times 3+1\times 4+\left( -\frac{1}{2} \right)\times 3 \\ \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, so it is the identity matrix Now, compute the matrix $B={{A}^{-1}}$ $B=\left[ \begin{matrix} \frac{7}{2} & -3 & \frac{1}{2} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ \end{matrix} \right]$ And $A=\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right]$ Determine the inverse of matrix A. Now, we will reduce matrix to row form: $A={{\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right]}^{-1}}$ Augment with an identity matrix, $A=\left[ \begin{matrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 1 & 3 & 4 & 0 & 1 & 0 \\ 1 & 4 & 3 & 0 & 0 & 1 \\ \end{matrix} \right]$ Swap matrix rows: \begin{align} & {{R}_{2}}\to {{R}_{2}}-1\times {{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-1\times {{R}_{1}} \\ & {{R}_{2}}\leftrightarrow {{R}_{3}} \\ & {{R}_{3}}\to {{R}_{3}}-\frac{1}{2}\times {{R}_{2}} \end{align} $A=\left[ \begin{matrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 2 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & -\frac{1}{2} & 1 & -\frac{1}{2} \\ \end{matrix} \right]$ Again \begin{align} & {{R}_{1}}\to {{R}_{1}}-3\times {{R}_{3}}] \\ & {{R}_{2}}\to \frac{1}{2}\times {{R}_{2}} \\ & {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{2}} \\ \end{align} $\left[ \begin{matrix} 1 & 0 & 0 & \frac{7}{2} & -3 & \frac{1}{2} \\ 0 & 1 & 0 & -\frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 0 & 1 & -\frac{1}{2} & 1 & -\frac{1}{2} \\ \end{matrix} \right]$ Thus, ${{A}^{-1}}=\left[ \begin{matrix} \frac{7}{2} & -3 & \frac{1}{2} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ \end{matrix} \right]$ Therefore, matrix $B$ is the multiplicative inverse of matrix $A$