## Precalculus (6th Edition) Blitzer

Matrix $B$ is a multiplicative inverse of the matrix $A$.
Find the product of $AB$ as follows: \begin{align} & AB=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & -7 \\ 0 & -1 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 7 \\ 0 & 1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( 1 \right)+0\left( 0 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( 4 \right)+0\left( 1 \right) & 1\left( 0 \right)+0\left( 7 \right)+0\left( 2 \right) \\ 0\left( 1 \right)+2\left( 0 \right)-7\left( 0 \right) & 0\left( 0 \right)+2\left( 4 \right)-7\left( 1 \right) & 0\left( 0 \right)+2\left( 7 \right)-7\left( 2 \right) \\ 0\left( 1 \right)-1\left( 0 \right)+4\left( 0 \right) & 0\left( 0 \right)-1\left( 4 \right)+4\left( 1 \right) & 0\left( 0 \right)-1\left( 7 \right)+4\left( 2 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 8-7 & 14-14 \\ 0 & -4+4 & -7+8 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \end{align} Therefore, $AB=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$ Next we will find the product of $BA$ as follows: \begin{align} & BA=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 7 \\ 0 & 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & -7 \\ 0 & -1 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\left( 1 \right)+0\left( 0 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( 2 \right)+0\left( -1 \right) & 1\left( 0 \right)+0\left( -7 \right)+0\left( 4 \right) \\ 0\left( 1 \right)+4\left( 0 \right)+7\left( 0 \right) & 0\left( 0 \right)+4\left( 2 \right)+7\left( -1 \right) & 0\left( 0 \right)+4\left( -7 \right)+7\left( 4 \right) \\ 0\left( 1 \right)+1\left( 0 \right)+2\left( 0 \right) & 0\left( 0 \right)+1\left( 2 \right)+2\left( -1 \right) & 0\left( 0 \right)+1\left( -7 \right)+2\left( 4 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 8-7 & 28-28 \\ 0 & 2-2 & -7+8 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \end{align} As $AB=BA=I$, where $I$ is an $3\times 3$ identity matrix, therefore, the given matrix $B$ is a multiplicative inverse of the matrix $A$.