# How do you multiply ((0, 1, 0), (2, -1, 1), (0, 2, -1)) with ((-1, 2, 0), (4, 6, 0), (1, 0, 1))?

Feb 18, 2016

Multiply $\text{row-line " xx " column-line of A"and } B$ to get C:
$C = \left(\begin{matrix}4 & 6 & 0 \\ - 5 & - 2 & 1 \\ 7 & 12 & - 1\end{matrix}\right)$

#### Explanation:

Let A = ((0,1,0 ), (2,-1,1 ), (0,2,−1 )) and B = ((−1,2,0 ), (4,6,0), (1,0,1 ))  then the product of $A \cdot B = C$ is given by:

$c i j = {a}_{i 1} {b}_{1 j} + {a}_{i 2} {b}_{2 j} + \cdots + {a}_{1 n} {b}_{1 n}$
$c i j = {\sum}_{k = 1}^{n} {a}_{i k} {b}_{k j}$
Now we have n = 3
$c i j = {\sum}_{k = \textcolor{red}{1}}^{\textcolor{red}{3}} {a}_{i k} {b}_{k j}$
c_(color(red)(1)color(blue)(1)) = a_(color(red)(1)1)b_(1color(blue)(1) +a_(color(red)(1)2)b_(2color(blue)(1) + a_(color(red)(1)3)b_(3color(blue)(1)

c_(color(red)(1)color(blue)(2)) = a_(color(red)(1)1)b_(1color(blue)(2) +a_(color(red)(1)2)b_(2color(blue)(2) + a_(color(red)(1)3)b_(3color(blue)(2)

$\cdots$

Note: each entry is made by three term obtained by multiplying row-line by column- line. That is:
row line = $\left(\left(\textcolor{red}{0} , \textcolor{b l u e}{1} , \textcolor{g r e e n}{0}\right)\right)$ $\left(\begin{matrix}\textcolor{red}{- 1} \\ \textcolor{b l u e}{2} \\ \textcolor{g r e e n}{0}\end{matrix}\right)$
$\textcolor{red}{\text{red"xx"red}}$ + $\textcolor{b l u e}{\text{blue"xx"blue}}$ + $\textcolor{g r e e n}{\text{green"xx"green}}$

Using this approach we find:

$C = \left(\begin{matrix}4 & 6 & 0 \\ - 5 & - 2 & 1 \\ 7 & 12 & - 1\end{matrix}\right)$