## Precalculus (6th Edition) Blitzer

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