Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 11

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 $