## Precalculus (6th Edition) Blitzer

$AB=BA={{I}_{3}}$
Consider the given matrices, \begin{align} & AB=\left[ \begin{array}{*{35}{r}} 1 & 2 & 2 \\ 2 & 3 & 3 \\ 1 & -1 & -2 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} -3 & 2 & 0 \\ 7 & -4 & 1 \\ -5 & 3 & -1 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -3+14-10 & 2-8+6 & 0+2-2 \\ -6+21-15 & 4-12+9 & 0+3-3 \\ -3-7+10 & 2+4-6 & 0-1+2 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] \end{align} Consider, \begin{align} & BA=\left[ \begin{array}{*{35}{r}} -3 & 2 & 0 \\ 7 & -4 & 1 \\ -5 & 3 & -1 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} 1 & 2 & 2 \\ 2 & 3 & 3 \\ 1 & -1 & -2 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -3+4+0 & -6+6+0 & -6+6+0 \\ 7-8+1 & 14-12-1 & 14-12-2 \\ -5+6-1 & -10+9+1 & -10+9+2 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] \end{align} Therefore $AB=BA={{I}_{3}}$ Hence, by the definition of the inverse, the matrix $B$ is the inverse of the matrix $A$.