Exercise 2.4.26

Answers

Columns of $A$ times rows of $B$ :

\begin{aligned} [\begin{matrix} 1 \\ 2 \\ 2 \end{matrix}] [\begin{matrix} 3 & 3 & 0 \end{matrix}] + [\begin{matrix} 0 \\ 4 \\ 1 \end{matrix}] [\begin{matrix} 1 & 2 & 1 \end{matrix}] = [\begin{matrix} 3 & 3 & 0 \\ 6 & 6 & 0 \\ 6 & 6 & 0 \end{matrix}] + [\begin{matrix} 0 & 0 & 0 \\ 4 & 8 & 4 \\ 1 & 2 & 1 \end{matrix}] = & [\begin{matrix} 3 & 3 & 0 \\ 10 & 14 & 4 \\ 7 & 8 & 1 \end{matrix}] = AB . \end{aligned}

mathchakra

2021-12-06 17:50