## Monday, December 19, 2005

### Review of Matrices

In today's blog, I will review some very basic results in 2x2 and 1x2 matrices.

This represents a very basic introduction that is meant to provide background for my larger blog on Fermat's Last Theorem: n = 5 (see here).

Today's blog is based on the work by Harold M. Stark in his book An Introduction to Number Theory.

1. Matrix defined

A matrix is a grouping of numbers that allows working on all the numbers at the same time.

For example, let's consider a 2 x 2 matrix that can be based on a set of numbers: 1, 2, 3, 4.

The matrix itself looks like this:

2. Addition and subtraction of matrices

Addition and subtraction of matrices are exactly the same as if you added and subtracted the numbers independently:

3. Multiplication of Numbers with Matrices

Multiplication with an integer just applies the integer to all the values involved so that:

4. Product of Two Matrices

In addition to these properites, matrices have there own special operations. The product of 2 matrices is a bit confusing. We define a product of a 1 x 2 matrix with a 2 x 2 matrix as the following:

We define a product a 2 x 2 matrix with a 2 x 2 matrix as the following:

Now, here's where it gets a bit confusing. We normally refer to a matrix using a capital letter. So let's say we have two matrices A,B such that: A is a 2x2 matrix and B is a 2x2 matrix. We cannot assume that AB = BA. For example, if we reverse the matrices above, we get the following equation:

Another important point is that there is no product defined for a 2x1 matrix and a 2x2 matrix or a 2x2 matrix and 1x2 matrix (since order is important in matrix products) and for that matter, there is no product defined a 2x2 matrix with a 1x2 matrix. In the case of 2x2 matrices, you can only get a product for a 2x2 matrix with a 2x2 matrix or a 1x2 matrix with a 2x2 matrix.

5. Determinant

A determinant is a value that is derived from a 2x2 matrix. Here is the definition:

Lemma 1: det(AB) = (detA)(detB)

(1) Let A =

Let B =

(2) AB =

(3) det(AB) = (ae+bg)(cf+dh) - (af+bh)(ce+dg) = (acef + adeh + bcfg + bdgh) - (acef + adfg + bceh + bdgh) = adeh + bcfg - adfg - bceh.

(4) det(A) = ad - bc
(5) det(B) = eh - fg
(6) So det(A)det(B) = (ad - bc)(eh - fg) = adeh + bcfg - adfg - bceh

QED

6. Identity Matrix

The Identity Matrix is referred to as I and defined as:

Lemma 2: AI = IA = A

(1) Let A =

(2) AI =

(3) IA =

QED

7. Inverse

Let A =

We denote the inverse of A as A-1 and we define it as:
A-1 =

Lemma 2: AA-1 = A-1A = I

(1)

(2)

QED

Lemma 3: det A-1 = 1/(det A)

(1) (det A)(det A-1) = det(AA-1) [From Lemma 1]

(2) det(AA-1) = det(I) [From Lemma 2]

(3) det(I) = 1*1 - 0*0 = 1. [Definition of I, Definition of Determinant]

(4) So, (det A)(det A-1) = 1

(5) And dividing both sides by (det A) gives us:
det A-1 = 1/(det A)

QED

7. Final Points

The last point here is that while AA-1 = I, it is not necessarily true that ABA-1 = B. The reason is that AB does not necessarily equal BA and we are not allowed to change the order of the matrix elements.

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Larry Freeman said...

Hi Mathcom,

I tried to contact you but was unable. Feel free to e-mail me directly at larry.freeman@gmail.com

I am very sorry that it took me so long to respond to this posting.

-Larry

Scouse Rob said...

In 5(6):
Should be:

Scouse Rob said...

In 7 (Final Points):
Should it be:
AB does not necessarily equal AB
AB does not necessarily equal A

Larry Freeman said...

Hi Rob,

Thanks again for the comments! I fixed both the typos that you found.

Cheers,

-Larry

aneesha RAGHUNATHAN said...

Thanks so much, this was very helpful!

Scouse Rob said...

In 4. Product of Two Matrices

Another important point is that there is no product defined for a 2x1 matrix and a 2x2 matrix or a 2x2 matrix and 2x1 matrix.

I thought that you could get the product of a 2x2 matrix and a 2x1 matrix (in that order) and that the result would be a 2x1 matrix.

Rob

Tina said...

Thanks so much, this math blog helps me refresh my memory of this than my professor did of explaining. :D thank you. :)

JonJun said...

Can you please verify what Scouse Rob said for his last comment?
I'm doing Hill-cipher question which C= KP mod26. KP (2x2 by 2x1) or PK(1x2 by 2x2)? Please help

Jon.

Larry Freeman said...

Hi JonJun,

Scouse Rob is right. That was a typo. I fixed it.

-Larry