# How do you multiply [(1,2,4), (-1,3,0)]*[(2,-4), (3,5), (-1,0)]?

Apr 14, 2017

Here is the method I was taught. (There is a dual method that is also useful.)

#### Explanation:

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

Find the first row of the product

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

$\left.\begin{matrix}1 \\ 2 \\ 4\end{matrix}\right. \left(\begin{matrix}2 & - 4 \\ 3 & 5 \\ - 1 & 0\end{matrix}\right)$

Now multiply times the first column and add to get the first number in the first row of the answer:
$1 \times 2 + 2 \times 3 + 4 \times - 1 = 2 + 6 + \left(- 4\right) = 4$

Next multiply times the second column and add to get the second number in the first row of the answer:
$1 \times - 4 + 2 \times 5 + 4 \times 0 = - 4 + 10 + 0 = 6$

(If there were more columns in the second matrix, we would continue this process.)

A this point we know that the product looks like:

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

Find the second row of the product

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

Make the second row vertical, multiply and add.

$\left.\begin{matrix}- 1 \\ 3 \\ 0\end{matrix}\right. \left(\begin{matrix}2 & - 4 \\ 3 & 5 \\ - 1 & 0\end{matrix}\right)$

gets us $- 1 \times 2 + 3 \times 3 + 0 \times \left(- 1\right) = - 2 + 9 + 0 = 7$

and $- 1 \times - 4 + 3 \times 5 + 0 \times 0 = 4 + 15 + 0 = 19$

So the product is:

$\left(\begin{matrix}1 & 2 & 4 \\ - 1 & 3 & 0\end{matrix}\right) \left(\begin{matrix}2 & - 4 \\ 3 & 5 \\ - 1 & 0\end{matrix}\right) = \left(\begin{matrix}4 & 6 \\ 7 & 19\end{matrix}\right)$

Note
If there were another row in the first matrix, then we would find a third row for the product.

Apr 14, 2017

This method is the dual of the one I used in the other answer.

#### Explanation:

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

Find the first column of the product

Take the first column of the second matrix. Use its entries to build a linear combination of the columns of the first matrix.
(We'll do the same for the second column in a minute.)

First column of second matrix: $\left(\begin{matrix}2 \\ 3 \\ - 1\end{matrix}\right)$

Linear combination of the columns of the first matrix:

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

this is the first column of the product.

A this point we know that the product looks like:

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

Find the second column of the product

Find the second column of the product by the same process using the values in second column of $\left(\begin{matrix}2 & - 4 \\ 3 & 5 \\ - 1 & 0\end{matrix}\right)$ as coefficients for a linear combination of the columns in the first matrix.

Second column of second matrix: $\left(\begin{matrix}- 4 \\ 5 \\ 0\end{matrix}\right)$

Linear combination of the columns of the first matrix:

$- 4 \left(\begin{matrix}1 \\ - 1\end{matrix}\right) + 5 \left(\begin{matrix}2 \\ 3\end{matrix}\right) + 0 \left(\begin{matrix}4 \\ 0\end{matrix}\right) = \left(\begin{matrix}- 4 + 10 + 0 \\ 4 + 15 + 0\end{matrix}\right) = \left(\begin{matrix}6 \\ 19\end{matrix}\right)$

this is the second column of the product.

So the product is:

$\left(\begin{matrix}1 & 2 & 4 \\ - 1 & 3 & 0\end{matrix}\right) \left(\begin{matrix}2 & - 4 \\ 3 & 5 \\ - 1 & 0\end{matrix}\right) = \left(\begin{matrix}4 & 6 \\ 7 & 19\end{matrix}\right)$