# How do you multiply ((2, -3), (5, 1)) by ((1, -3, 4), (2, 4, 3))?

Aug 18, 2016

$\left(\begin{matrix}2 & - 3 \\ 5 & 1\end{matrix}\right) \times \left(\begin{matrix}1 & - 3 & 4 \\ 2 & 4 & 3\end{matrix}\right) = \left(\begin{matrix}- 4 & - 18 & - 1 \\ 7 & - 11 & 23\end{matrix}\right)$

#### Explanation:

Check that the matrices are compatible.
Look at the format of each matrix - the Rows (R) and the Columns
(C)
R x C

The two 'middle' numbers must be the $\textcolor{red}{\text{same}}$
$\textcolor{b l u e}{2} \times \textcolor{red}{2} \mathmr{and} \textcolor{red}{2} \times \textcolor{b l u e}{3}$

These are compatible. The answer will be a $\textcolor{b l u e}{2 \times 3}$ matrix. $\left(\begin{matrix}{R}_{1} {C}_{1} & {R}_{1} {C}_{2} & {R}_{1} {C}_{3} \\ {R}_{2} {C}_{1} & {R}_{2} {C}_{2} & {R}_{2} {C}_{3}\end{matrix}\right)$

Each row in matrix 1 must be multiplied by each column in matrix 2.
The elements are all multiplied and then added together.

$\left(\begin{matrix}2 & - 3 \\ 5 & 1\end{matrix}\right)$ by $\left(\begin{matrix}1 & - 3 & 4 \\ 2 & 4 & 3\end{matrix}\right) = \left(\begin{matrix}- 4 & - 18 & - 1 \\ 7 & - 11 & 23\end{matrix}\right)$

${R}_{1} {C}_{1} = 2 - 6 = - 4$
${R}_{1} {C}_{2} = - 6 - 12 = - 18$
${R}_{1} {C}_{3} = 8 - 9 = - 1$
${R}_{2} {C}_{1} = 5 + 2 = 7$
${R}_{2} {C}_{2} = - 15 + 4 = - 11$
${R}_{2} {C}_{3} = 20 + 3 = 23$