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Multiplying Matrices I

Module by: Kenny M. Felder. E-mail the author

Summary: This module provides sample problems which develop concepts related to multiplying matrices.

Just to limber up your matrix muscles, let’s try doing the following matrix addition.

Exercise 1

[ 2 5 x 3 7 2y ] + [ 2 5 x 3 7 2y ] + [ 2 5 x 3 7 2y ] = [ 2 5 x 3 7 2y ]+[ 2 5 x 3 7 2y ]+[ 2 5 x 3 7 2y ]=

Exercise 2

How many times did you add that matrix to itself?

Exercise 3

Rewrite #1 as a multiplication problem. (*Remember what multiplication means—adding something to itself a bunch of times!)

This brings us to the world of multiplying a matrix by a number. It’s very straightforward. You end up with a matrix that has the same dimensions as the original, but all the individual cells have been multiplied by that number.

Let’s do another example. I’m sure you remember Professor Snape’s grade matrix.

Table 1
  Poison Cure Love philter Invulnerability
Granger, H 100 105 99 100
Longbottom, N 80 90 85 85
Malfoy, D 95 90 0 85
Potter, H 70 75 70 75
Weasley, R 85 90 95 90

Now, we saw how Professor Snape could lower his grades (which he loves to do) by subtracting a curve matrix. But there is another way he can lower his grades, which is by multiplying the entire matrix by a number. In this case, he is going to multiply his grade matrix by 910910 size 12{ { {9} over {"10"} } } {}. If we designate his grade matrix as [S][S] then the resulting matrix could be written as 910910 size 12{ { {9} over {"10"} } } {}[S][S].

Note:

Remember that the cells in a matrix are numbers! So [S][S] is just the grades, not the names.

Exercise 4

Write down the matrix 910910 size 12{ { {9} over {"10"} } } {}[S][S].

Finally, it’s time to Prof. Snape to calculate final grades. He does this according to the following formula: “Poison” counts 30%, “Cure” counts 20%, “Love philter” counts 15%, and the big final project on “Invulnerability” counts 35%. For instance, to calculate the final grade for “Granger, H” he does the following calculation: ( 30 % ) ( 100 ) + ( 20 % ) ( 105 ) + ( 15 % ) ( 99 ) + ( 35 % ) ( 100 ) = 100.85 (30%)(100)+(20%)(105)+(15%)(99)+(35%)(100)=100.85.

To make the calculations easier to keep track of, the Professor represents the various weights in his grading matrix which looks like the following:

. 3 . 2 . 15 . 35 . 3 . 2 . 15 . 35 size 12{ left [ matrix { "." 3 {} ## "." 2 {} ## "." "15" {} ## "." "35" } right ]} {}

The above calculation can be written very concisely as multiplying a row matrix by a column matrix, as follows.

1001059910010010599100 size 12{ left [ matrix { "100" {} # "105" {} # "99" {} # "100"{} } right ]} {}.3.2.15.35 = [ 100.85 ] .3.2.15.35 size 12{ left [ matrix { "." 3 {} ## "." 2 {} ## "." "15" {} ## "." "35" } right ]} {}=[100.85]

A “row matrix” means a matrix that is just one row. A “column matrix” means…well, you get the idea. When a row matrix and a column matrix have the same number of items, you can multiply the two matrices. What you do is, you multiply both of the first numbers, and you multiply both of the second numbers, and so on…and you add all those numbers to get one big number. The final answer is not just a number—it is a 1×1 matrix, with that one big number inside it.

Exercise 5

Below, write the matrix multiplication that Prof. Snape would do to find the grade for “Potter, H”. Show both the problem (the two matrices being multiplied) and the answer (the 1×1 matrix that contains the final grade).

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