# How do you do you multiply a 3x2 matrix by a 2x2 matrix??

Oct 22, 2015

Here is an example (rather than a formula, it is a process).

#### Explanation:

Here is an example

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

Find the first row of the product

Take the first row of $\left(\begin{matrix}4 & 5 \\ 3 & 4 \\ 1 & 2\end{matrix}\right)$, and make it vertical. (We'll do the same for the second row in a minute. And then for the third row.)

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

Now multiply times the first column and add to get the first number in the first row of the answer:
$4 \times 3 + 5 \times 0 = 12 + 0 = 12$

Next multiply times the second column and add to get the second number in the first row of the answer:
$4 \times 1 + 5 \times 3 = 4 + 15 = 19$

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

A this point we know that the product looks like:

$\left(\begin{matrix}4 & 5 \\ 3 & 4 \\ 1 & 2\end{matrix}\right) \left(\begin{matrix}3 & 1 \\ 0 & 3\end{matrix}\right) = \left(\begin{matrix}12 & 19 \\ \text{-" & "-" \\ "-" & "-}\end{matrix}\right)$

Find the second row of the product

Find the second row of the product by the same process using the second row of $\left(\begin{matrix}4 & 5 \\ 3 & 4 \\ 1 & 2\end{matrix}\right)$

Make the second row veritcal, muliptly and add.

$\left.\begin{matrix}3 \\ 4\end{matrix}\right. \left(\begin{matrix}3 & 1 \\ 0 & 3\end{matrix}\right)$ gets us $3 \times 3 + 4 \times 0 = 9$ and $3 \times 1 + 4 \times 3 = 15$

A this point we know that the product looks like:

$\left(\begin{matrix}4 & 5 \\ 3 & 4 \\ 1 & 2\end{matrix}\right) \left(\begin{matrix}3 & 1 \\ 0 & 3\end{matrix}\right) = \left(\begin{matrix}12 & 19 \\ 9 & 15 \\ \text{-" & "-}\end{matrix}\right)$

Find the third row of the product.

$\left.\begin{matrix}1 \\ 2\end{matrix}\right. \left(\begin{matrix}3 & 1 \\ 0 & 3\end{matrix}\right)$ to get: $3 + 0 = 3$ and $1 + 6 = 7$

$\left(\begin{matrix}4 & 5 \\ 3 & 4 \\ 1 & 2\end{matrix}\right) \left(\begin{matrix}3 & 1 \\ 0 & 3\end{matrix}\right) = \left(\begin{matrix}12 & 19 \\ 9 & 15 \\ 3 & 7\end{matrix}\right)$