3 Some surprising results

We have already calculated the product A B where

A = 1 2 3 4 and B = 1 1 2 1

Now complete the following task in which you are asked to determine the product B A , i.e. with the matrices in reverse order.

Task!

For matrices A = 1 2 3 4 and B = 1 1 2 1 form the products of

row 1 of B and column 1 of A row 1 of B and column 2 of A

row 2 of B and column 1 of A row 2 of B and column 2 of A

Now write down the matrix B A :

row 1, column 1 is 1 × 1 + ( 1 ) × 3 = 2 row 1, column 2 is 1 × 2 + ( 1 ) × 4 = 2

row 2, column 1 is 2 × 1 + 1 × 3 = 1   row 2, column 2 is 2 × 2 + 1 × 4 = 0

B A is 2 2 1 0

It is clear that A B and B A are not in general the same. In fact it is the exception that A B = B A .

In the special case in which A B = B A we say that the matrices A and B commute .

Task!

Calculate A B and B A where

A = a b c d and B = 0 0 0 0

A B = B A = 0 0 0 0

We call B the 2 × 2 zero matrix written 0 so that A × 0 ̲ = 0 ̲ × A = 0 for any matrix A .

Now in the multiplication of numbers, the equation

a b = 0

implies that either a is zero or b is zero or both are zero. The following task shows that this is not necessarily true for matrices.

Task!

Carry out the multiplication A B where

A = 1 1 1 1 , B = 1 1 1 1

A B = 0 0 0 0

Here we have a zero product yet neither A nor B is the zero matrix! Thus the statement A B = 0 does not allow us to conclude that either A = 0 ̲ or B = 0 ̲ .

Task!

Find the product A B where A = a b c d and B = 1 0 0 1

A B = a b c d = A

The matrix 1 0 0 1 is called the identity matrix or unit matrix of order 2 , and is usually denoted by the symbol I . (Strictly we should write I 2 , to indicate the size.) I plays the same role in matrix multiplication as the number 1 does in number multiplication.

Hence

just as a × 1 = 1 × a = a for any number a , so A I = I A = A for any matrix A .