# How do you multiply ((0, 6), (5, -5)) and ((1, 1, 1), (0, 0, -2))?

Feb 15, 2016

${C}_{\textcolor{b l u e}{23}} = \left(\begin{matrix}0 & 6 \\ 5 & - 5\end{matrix}\right) \cdot \left(\begin{matrix}1 & 1 & 1 \\ 0 & 0 & - 2\end{matrix}\right) = \left(\begin{matrix}0 & 0 & - 12 \\ 5 & 5 & 15\end{matrix}\right)$

#### Explanation:

to multiply a matrix by another matrix we need the number of columns to match the number of rows.
${C}_{i k} = {A}_{i \textcolor{red}{1}} {B}_{\textcolor{red}{1} k} - {A}_{i \textcolor{red}{2}} {B}_{\textcolor{red}{2} k} + \cdots + {A}_{i \textcolor{red}{N}} {B}_{\textcolor{red}{N} k}$
note the inner counter, $\textcolor{red}{j}$ is the same that is the column of A is equal to the row of B ...
${C}_{i k} = {\sum}_{j = 1}^{N} {A}_{i \textcolor{red}{j}} {B}_{\textcolor{red}{j} k}$
$\left(\begin{matrix}{c}_{11} & {c}_{12} & \cdots & {c}_{1 k} \\ \vdots & \vdots & \vdots & \vdots \\ {c}_{k 1} & {c}_{k 2} & \cdots & {c}_{k k}\end{matrix}\right) = \left(\begin{matrix}{a}_{11} & {a}_{12} & \cdots & {a}_{1 j} \\ \vdots & \vdots & \vdots & \vdots \\ {a}_{i 1} & {a}_{i 2} & \cdots & {a}_{i j}\end{matrix}\right) \cdot \left(\begin{matrix}{b}_{11} & {b}_{12} & \cdots & {b}_{1 k} \\ \vdots & \vdots & \vdots & \vdots \\ {b}_{j 1} & {b}_{j 2} & \cdots & {b}_{j k}\end{matrix}\right)$

Now that we had a quick review on Matrix multiplication in you case: A_(2color(red)2) = ((0,6),(5,-5)); B_(color(red)(2)3)=((1,1,1),(0,0,-2))
So the multiplication order is:
C_color(blue)(23) = A_(color(blue)2color(red)2)*B_(color(red)(2)color(blue)(3)
${C}_{\textcolor{b l u e}{23}} = \left(\begin{matrix}0 & 6 \\ 5 & - 5\end{matrix}\right) \cdot \left(\begin{matrix}1 & 1 & 1 \\ 0 & 0 & - 2\end{matrix}\right) = \left(\begin{matrix}0 & 0 & - 12 \\ 5 & 5 & 15\end{matrix}\right)$