Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 7

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 $.