## Precalculus (6th Edition) Blitzer

a) $AB=\left[ \begin{matrix} -1 & -2 & -3 \\ -2 & -4 & -6 \\ -3 & -6 & -9 \\ \end{matrix} \right]$ b) $BA=\left[ -14 \right]$
(a) Consider the product, \begin{align} & AB=\left[ \begin{matrix} -1 \\ -2 \\ -3 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1\left( 1 \right) & -1\left( 2 \right) & -1\left( 3 \right) \\ -2\left( 1 \right) & -2\left( 2 \right) & -2\left( 3 \right) \\ -3\left( 1 \right) & -3\left( 2 \right) & -3\left( 3 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & -2 & -3 \\ -2 & -4 & -6 \\ -3 & -6 & -9 \\ \end{matrix} \right] \end{align} So that, $AB=\left[ \begin{matrix} -1 & -2 & -3 \\ -2 & -4 & -6 \\ -3 & -6 & -9 \\ \end{matrix} \right]$ (b) Consider the product, \begin{align} & BA=\left[ \begin{matrix} 1 & 2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} -1 \\ -2 \\ -3 \\ \end{matrix} \right] \\ & =\left[ 1\left( -1 \right)+2\left( -2 \right)+3\left( -3 \right) \right] \\ & =\left[ -1-4-9 \right] \\ & =\left[ -14 \right] \end{align}