# How do you multiply (4, 5)(3, 4) by (3, 0)(0, 3)?

Jul 4, 2015

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

#### Explanation:

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

Find the first row of the product

Take the first row of $\left(\begin{matrix}4 & 5 \\ 3 & 4\end{matrix}\right)$, and make it vertical in front of $\left(\begin{matrix}3 & 0 \\ 0 & 3\end{matrix}\right)$. (We'll do the same for the second row in a minute.)

It looks like:

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

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:
$4 \times 0 + 5 \times 3 = 0 + 15 = 15$

The first row of the product is: $\left(\left(12 , 15\right)\right)$

A this point we know that the product looks like:

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

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

$\left.\begin{matrix}3 \\ 4\end{matrix}\right. \left(\begin{matrix}3 & 0 \\ 0 & 3\end{matrix}\right)$ to get: $9$ and $12$

The second row of the product is: $\left(\left(9 , 12\right)\right)$

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

Jul 4, 2015

The second matrix is just a scalar multiple ($3$) of the identity matrix, so the result is just the same scalar multiple of the first matrix, i.e.

$3 \left(\begin{matrix}4 & 5 \\ 3 & 4\end{matrix}\right) = \left(\begin{matrix}12 & 15 \\ 9 & 12\end{matrix}\right)$

#### Explanation:

Jim's explanation is really helpful if you are multiplying any two square matrices, but in this particular example there's a short cut.

$\left(\begin{matrix}3 & 0 \\ 0 & 3\end{matrix}\right) = 3 \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

So:

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

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

$= 3 \left(\begin{matrix}4 & 5 \\ 3 & 4\end{matrix}\right)$

$= \left(\begin{matrix}12 & 15 \\ 9 & 12\end{matrix}\right)$