The tutor demonstrates the technique of multiplying matrices.

To go further with Markov chains (introduced in my previous post), the reader needs to understand matrix multiplication. To many, the method is surprising at first.

Example: Consider the matrices A and B. Find the matrix product A x B:

In preparation, the reader needs to know that matrix entries are commonly referred to by (row,column). Therefore, entry (1,3) in matrix A, above, is 9.

Solution:

We start by mulitplying row one of A by column one of B, as follows:

-1(9)+5(2)+9(0)=1

This result is entry (1,1) of the solution matrix S.

Next, we multiply the first row of A by the second column of B to get entry (1,2) of S:

-1(0)+5(1)+9(-7)=-58 = S(1,2)

We move on to the second row of A, multiplying it by the first column of B. The result will be (2,1) of S:

2(9)+0(2)+3(0)=18 = S(2,1)

Finally, we multiply the second row of A by the second column of B, yielding (2,2) in S:

2(0)+0(1)+3(-7)=-21 = S(2,2)

The solution matrix, S=AxB, appears as follows:

I’ll continue about this topic in coming posts.

HTH:)

Source:

Johnson/Riess/Arnold. Introduction to Linear Algebra. Don Mills: Addison-Wesley   Publishing Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.