# Multiplying Matrices

Go forth and multiply... matrices.

 Algebra II Matrices Language English Language

### Transcript

00:29

per uniform.

00:31

What is the total cost of equipment for each team?

00:34

Well, we know that to find the total cost for each team, we'll need to multiply the

00:39

number of equipment by the cost.

00:42

Remember, that if we're going to perform operations with our data, we should put our data in matrix

00:47

format...with equipment numbers in one matrix and costs in another matrix. [Matrices of the values]

00:52

But before we can multiply two matrices, we need to make sure that the number of columns

00:57

in matrix A is the same as the number of rows in matrix B.

01:02

In this case, we have 3 columns in the first matrix, and 3 rows in the second one.

01:06

Checks out!

01:07

To multiply the two, we'll be doing something funky.

01:11

We'll multiply corresponding entries in the first row of matrix A and the first column

01:15

01:18

So...the first entry in the matrix is 5 times 10 plus 20 times 2 plus 15 times 20. [Finger points to the entries in the matrix]

01:25

And the second entry is 6 times 10 plus 15 times 2 plus 14 times 20.

01:31

5 times 10 is 50, 20 times 2 is 40, and 15 times 20 is 300.

01:36

So 50 plus 40 plus 300 is 390, which gives the men's team total cost of \$390 for their

01:43

equipment.

01:44

For the women's team...we have that 6 times 10 is 60; 15 times 2 is 30 and 14 times 20 [Total cost for the mens team written in a matrix]

01:49

is 280.

01:50

60 plus 30 plus 280 is 370...so the women's team total cost of equipment is 370 bucks. [The cost of the women's team is added to the matrix]

01:58

Notice that our ending matrix has dimensions of 2 by 1, which is also the row of matrix

02:05

A and column of matrix B. If he keeps up the good work Jerry may soon be

02:10

promoted to errand man... [Jerry cries and team mate rubs him on the back]