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

^{-1}A = 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.

## 10 comments :

hi

we have a blog on math and want to have relationship with you. contact us and join.

we publish our blog in english/french/persian language.

join us and contact me...

http://mathcom.blogfa.com

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

In 5(6):

(ad - bc)(eh - fg) = adeh + bcfg - adfh - bceh

Should be:

(ad - bc)(eh - fg) = adeh + bcfg - adfg - bceh

In 7 (Final Points):

Should it be:

AB does not necessarily equal AB

Instead of:

AB does not necessarily equal A

Hi Rob,

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

Cheers,

-Larry

Thanks so much, this was very helpful!

In

4. Product of Two MatricesAnother 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

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

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.

Hi JonJun,

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

-Larry

Post a Comment