# How do you multiply ((0, 1, 0), (6, -3, 1), (-1, 4, -2))) and ((6, 7, 1), (2, 10, 5), (1, -10, 9))?

Nov 19, 2016

The answer is $= \left(\begin{matrix}2 & 10 & 5 \\ 31 & 2 & 0 \\ 0 & 53 & 1\end{matrix}\right)$

#### Explanation:

The multiplication of 2 matrices is

$\left(\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}\right) \cdot \left(\begin{matrix}k & l & m \\ n & p & q \\ r & s & t\end{matrix}\right)$

$\left(\begin{matrix}a k + b n + c r & a l + b p + c s & a m + b q + c t \\ \mathrm{dk} + + e n + f r & \mathrm{dl} + e p + f s & \mathrm{dm} + e q + f t \\ g k + h n + i r & g l + h p + i s & g m + h q + i t\end{matrix}\right)$

$\left(\begin{matrix}0 & 1 & 0 \\ 6 & - 3 & 1 \\ - 1 & 4 & - 2\end{matrix}\right) \cdot \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right)$

$= \left(\begin{matrix}2 & 10 & 5 \\ 31 & 2 & 0 \\ 0 & 53 & 1\end{matrix}\right)$

Nov 19, 2016

There is a formula, but many find the following process easier to remember.

#### Explanation:

$\left(\begin{matrix}0 & 1 & 0 \\ 6 & - 3 & 1 \\ - 1 & 4 & - 2\end{matrix}\right) \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right)$

Find the first row of the product

Take the first row of $\left(\begin{matrix}0 & 1 & 0 \\ 6 & - 3 & 1 \\ - 1 & 4 & - 2\end{matrix}\right)$, and make it vertical. (We'll do the same for the second row in a minute.)

$\left.\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right. \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right)$

Now multiply times the first column and add to get the first number in the first row of the answer:
$\left(0 \times 6\right) + \left(1 \times 2\right) + \left(0 \times 1\right) = 0 + 2 + 0 = 2$

Next multiply times the second column and add to get the second number in the first row of the answer:
$\left(0 \times 7\right) + \left(1 \times 10\right) + \left(0 \times - 10\right) = 0 + 10 + 0 = 10$

Finally, multiply times the third column and add to get the third number in the first row of the answer:
$\left(0 \times 1\right) + \left(1 \times 5\right) + \left(0 \times 9\right) = 0 + 5 + 0 = 5$

At this point we know that the product looks like:

$\left(\begin{matrix}0 & 1 & 0 \\ 6 & - 3 & 1 \\ - 1 & 4 & - 2\end{matrix}\right) \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right) = \left(\begin{matrix}2 & 10 & 5 \\ \text{-" & "-" & "-" \\ "-" & "-" & "-}\end{matrix}\right)$

Find the second row of the product
Find the second row of the product by the same process using the second row of $\left(\begin{matrix}0 & 1 & 0 \\ 6 & - 3 & 1 \\ - 1 & 4 & - 2\end{matrix}\right)$

Stand up the second row, then multiply and add to find the numbers in the second row of the answer.
$\left.\begin{matrix}6 \\ - 3 \\ 1\end{matrix}\right. \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right)$

First number in the second row of the answer:
$\left(6 \times 6\right) + \left(- 3 \times 2\right) + \left(1 \times 1\right) = 36 - 6 + 1 = 31$

Second number in the second row of the answer:
$\left(6 \times 7\right) + \left(- 3 \times 10\right) + \left(1 \times - 10\right) = 42 - 30 - 10 = 2$

Third number in the second row of the answer:
$\left(6 \times 1\right) + \left(- 3 \times 5\right) + \left(1 \times 9\right) = 6 - 15 + 9 = 0$

At this point we know that the product looks like:

$\left(\begin{matrix}0 & 1 & 0 \\ 6 & - 3 & 1 \\ - 1 & 4 & - 2\end{matrix}\right) \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right) = \left(\begin{matrix}2 & 10 & 5 \\ 31 & 2 & 0 \\ \text{-" & "-" & "-}\end{matrix}\right)$

Find the third row of the product.

$\left.\begin{matrix}- 1 \\ 4 \\ - 2\end{matrix}\right. \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right)$ to get:

$- 6 + 8 - 2 = 0$, then $- 7 + 40 + 20 = 53$, then $- 1 + 20 - 18 = 1$

$\left(\begin{matrix}0 & 1 & 0 \\ 6 & - 3 & 1 \\ - 1 & 4 & - 2\end{matrix}\right) \left(\begin{matrix}6 & 7 & 1 \\ 2 & 10 & 5 \\ 1 & - 10 & 9\end{matrix}\right) = \left(\begin{matrix}2 & 10 & 5 \\ 31 & 2 & 0 \\ 0 & 53 & 1\end{matrix}\right)$