Precalculus (6th Edition) Blitzer

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