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Teachers & SchoolsGo forth and multiply... matrices.

Algebra II | Matrices |

Language | English Language |

per uniform.

What is the total cost of equipment for each team?

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

number of equipment by the cost.

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

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

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

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

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

Checks out!

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

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

of B...then add.

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]

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

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

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

equipment.

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]

is 280.

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]

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

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

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