Problem:
Multiply
$\left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right)$ and
$\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)$.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Multiply the following matrices}: \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right) \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 
\begin{array}{l}
 \text{The }\text{dimensions }\text{of }\text{the }\text{first }\text{matrix }\text{are }3\times 2 \text{and }\text{the }\text{dimensions }\text{of }\text{the }\text{second }\text{matrix }\text{are }2\times 1. \\
 \text{This }\text{means }\text{the }\text{dimensions }\text{of }\text{the }\text{product }\text{are }3\times 1: \\
\end{array}
 \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 \_ \\
 \_ \\
 \_ \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Since }\text{both }\text{matrices }\text{are }\text{lower }\text{triangular, }\text{the }\text{product }\text{will }\text{also }\text{be }\text{lower }\text{triangular. }\text{This }\text{means }\text{we }\text{can }\text{place }0\text{'s} \text{above }\text{the }\text{main }\text{diagonal}: \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 \_ \\
 \_ \\
 \_ \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Highlight }\text{the }1^{\text{st}} \text{row }\text{and }\text{the }1^{\text{st}} \text{column}: \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 \_ \\
 \_ \\
 \_ \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{l}
 \text{Multiply }\text{corresponding }\text{components }\text{and }\text{add: }0\ 2+0 (-1)=0. \\
 \text{Place }\text{this }\text{number }\text{into }\text{the }1^{\text{st}} \text{row }\text{and }1^{\text{st}} \text{column }\text{of }\text{the }\text{product}: \\
\end{array}
 \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 \fbox{$0$} \\
 \_ \\
 \_ \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Highlight }\text{the }2^{\text{nd}} \text{row }\text{and }\text{the }1^{\text{st}} \text{column}: \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 \_ \\
 \_ \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{l}
 \text{Multiply }\text{corresponding }\text{components }\text{and }\text{add: }2\ 2+1 (-1)=3. \\
 \text{Place }\text{this }\text{number }\text{into }\text{the }2^{\text{nd}} \text{row }\text{and }1^{\text{st}} \text{column }\text{of }\text{the }\text{product}: \\
\end{array}
 \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 \fbox{$3$} \\
 \_ \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Highlight }\text{the }3^{\text{rd}} \text{row }\text{and }\text{the }1^{\text{st}} \text{column}: \\
 \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 3 \\
 \_ \\
\end{array}
\right) \\
\end{array}
 \\
 
\begin{array}{l}
 
\begin{array}{l}
 \text{Multiply }\text{corresponding }\text{components }\text{and }\text{add: }0\ 2+(-1)\, (-1)=1. \\
 \text{Place }\text{this }\text{number }\text{into }\text{the }3^{\text{rd}} \text{row }\text{and }1^{\text{st}} \text{column }\text{of }\text{the }\text{product}: \\
\end{array}
 \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & \left(
\begin{array}{cc}
 0 & 0 \\
 2 & 1 \\
 0 & -1 \\
\end{array}
\right).\left(
\begin{array}{c}
 2 \\
 -1 \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 3 \\
 \fbox{$1$} \\
\end{array}
\right) \\
\end{array}
$} \\
\end{array}
 \\
\end{array}
{'source': 'AMPS', 'file': 'amps/mathematica/linear_algebra/multiply_w_steps/3821.txt'}