Does the commutative property hold for the product of matrices?

Question

Does A X B = B X A in all cases? If not, please show a counter example.

Answer 1

Matrix multiplication is not commutative; below is a counterexample.

Let A =
[1 2 ]
[3 0 ]

B =
[0 1]
[-1 1]

AB =
[1 2 ] [0 1]
[3 0 ] [-1 1]

= [-2 3]
= [0 3]

BA =
[0 1] [1 2]
[-1 1] [3 0]

= [3 0]
= [2 -2]

Notice that AB does NOT equal BA.

Haroun: It is not true that only matrices and their inverse commute. Consider this example:

A =
[2 1]
[1 1]

B =
[3 -1]
[-1 4]

AB = BA =
[5 2]
[2 3]

In this case, A and B are not inverses of each other, but they still commute.

If A and B are both square matrices and one of A or B is the identity matrix, they also commute.

A =
[a b]
[c d]

B =
[1 0]
[0 1]

AB = BA =
[a b]
[c d]

Answer 2

No. It does in some cases, but not always. Here are two examples:

1) If the matrices are not square then you either cannot multiply them in the opposite order, or if you do you will get a matrix of a different size. Example:

| 1 | * | 1 2 | = | 5 |
| 2 |

but multiplying it in reverse order gives you a 2x2 matrix.

2) Also not all pairs of square matrices with the same dimensions commute. For example, for the product:

| 1 2 | * | 3 4 |
| 3 4 | | 1 2 |

the result is not the same if you reverse the order.

However, note that a size nxn identity matrix (all 1's on the main diagonal, all 0's elsewhere) commutes with every other nxn matrix.

Hope that helps!

Answer 3

No. It is particularly obvious for non-square matrices, but even square ones do not necessarily commute. I have not tried this, but see what happens when you separately pre-multiply and post-multiply a small arbitrary square matrix by the identity matrix.

Answer 4

No they don't commute. And somebody gave you a counter example. On a sidenote, only a matrix and its inverse commute.

Answer 5

no tyhiis is basic stuff when you encouter matrices in class