# Matrix multiplication Question

• August 9th 2010, 09:09 PM
jsridhar72
Matrix multiplication Question
I have 2 matrices(2x2) say A,B. I want to get a 3rd matrix C by ABA(multiplying B on both sides.,that is left and right sides of B with the matrix A). Is there any special property involved in multiplying one matrix B with another matrix A on both sides. What are the properties of the resultant matrix C. I need to understand this concept for a presentation. Please help
• August 9th 2010, 09:13 PM
pickslides
Matrices don't commute which means AB may not equal BA.

The only other property you need to know is that matrices can only be multiplied if the order of the matrices allow it. As you have 3 matrices that are 2x2 it is possible to get a result ABA.

Does this make sense?
• August 9th 2010, 09:23 PM
jsridhar72
Quote:

Originally Posted by pickslides
Matrices don't commute which means AB may not equal BA.

The only other property you need to know is that matrices can only be multiplied if the order of the matrices allow it. As you have 3 matrices that are 2x2 it is possible to get a result ABA.

Does this make sense?

The matries A,B are taken in such a way that they wont commute. That is AB not Equals BA. more clarification please.
• August 9th 2010, 09:30 PM
pickslides
Using real numbers...

Addition commutes i.e $3+5 = 5+3$

Subtraction does not i.e $3-5 \neq 5-3$

but matrices don't work like this. Have you been shown how to multiply matrices?
• August 9th 2010, 09:47 PM
jsridhar72
Quote:

Originally Posted by pickslides
Using real numbers...

Addition commutes i.e $3+5 = 5+3$

Subtraction does not i.e $3-5 \neq 5-3$

but matrices don't work like this. Have you been shown how to multiply matrices?

Thank you any way for your reply...I know matrix multiplication...But one thing I forget to tell is that the matrices A,B are taken from a group GL(r,Zn). Every element in the matrix is multiplicative modulo n where n is the product of two large prime numbers. Is this makes difference?
• August 9th 2010, 09:53 PM
pickslides
Quote:

Originally Posted by jsridhar72
Thank you any way for your reply...I know matrix multiplication...But one thing I forget to tell is that the matrices A,B are taken from a group GL(r,Zn). Every element in the matrix is multiplicative modulo n where n is the product of two large prime numbers. Is this makes difference?

This is a big detail to leave out! I'm sure it might make a difference.
• August 9th 2010, 10:04 PM
jsridhar72
Quote:

Originally Posted by pickslides
This is a big detail to leave out! I'm sure it might make a difference.

Thanks...Though I am not able to get what I want I will come back with something else soon...Thank for your patience reply and you people are great value for people like me who treat maths as aliens...
• August 10th 2010, 02:22 AM
Vlasev
I think you mean that the entries are $A,B \in GL(2,\mathbb{Z}_n)$?

You probably know this but I will say it anyway to double check. $GL(r,\mathbb{Z}_n)$ is the general linear group over the cyclic group $\mathbb{Z}_n$ (field). It has dimension $r^2$ and is a subset of the $r \times r$ matrices. It consists of the invertible matrices, so $A$ and $B$ are invertible matrices. That means that each determinant is a unit in $\mathbb{Z}_n$.
• August 10th 2010, 03:22 AM
jsridhar72
Quote:

Originally Posted by Vlasev
I think you mean that the entries are $A,B \in GL(2,\mathbb{Z}_n)$?

You probably know this but I will say it anyway to double check. $GL(r,\mathbb{Z}_n)$ is the general linear group over the cyclic group $\mathbb{Z}_n$ (field). It has dimension $r^2$ and is a subset of the $r \times r$ matrices. It consists of the invertible matrices, so $A$ and $B$ are invertible matrices. That means that each determinant is a unit in $\mathbb{Z}_n$.

Thank you sir.What you said is true. If AB is invertible what is C=ABA? Thank you in advance
• August 10th 2010, 03:36 AM
Vlasev
Well if two matrices A and B are each invertible, so is their product. Thus ABA is invertible and hence C is invertible. The determinant of C is a unit in $\mathbb{Z}_n$. I don't know what more to say than this.
• August 10th 2010, 09:24 PM
jsridhar72
Quote:

Originally Posted by Vlasev
Well if two matrices A and B are each invertible, so is their product. Thus ABA is invertible and hence C is invertible. The determinant of C is a unit in $\mathbb{Z}_n$. I don't know what more to say than this.

Thank you very much sir...I am not great in maths...Even fundamental things...The jargons are mind boggling...I Studied computer science and no maths except discrete maths....I will ask explanations to many jargons which you may feel very simple...My first jargon...What is isomorphism? Particularly isomorphism in groups?Thanks in advance.
• August 10th 2010, 10:13 PM
Vlasev
An isomorphism is a function that takes elements from the set (group in this case) and maps them to other elements in the group. It must be a bijection (one to one and onto, that is, it maps exactly one element to exactly one element and it doesn't map two different elements to the same one).

i.e.

a ------- b
b ------- c
c ------- a
d ------- d

is one such bijection.

a ------- a
b ------- a
c ------- d
d ------- c

is not a bijectino because it maps two elements onto the same one.

However, this is not all. The isomorphism is a special bijection in the sense that for F to be an isomorphism from G to G it has to also satisfy the following property:

F(a.b) = F(a).F(b)

where the dot is the group operation, which in your case is matrix multiplication.
• August 10th 2010, 10:31 PM
jsridhar72
Quote:

Originally Posted by Vlasev
An isomorphism is a function that takes elements from the set (group in this case) and maps them to other elements in the group. It must be a bijection (one to one and onto, that is, it maps exactly one element to exactly one element and it doesn't map two different elements to the same one).

i.e.

a ------- b
b ------- c
c ------- a
d ------- d

is one such bijection.

a ------- a
b ------- a
c ------- d
d ------- c

is not a bijectino because it maps two elements onto the same one.

However, this is not all. The isomorphism is a special bijection in the sense that for F to be an isomorphism from G to G it has to also satisfy the following property:

F(a.b) = F(a).F(b)

where the dot is the group operation, which in your case is matrix multiplication.

Hmm...hmmm..Excellent sir...I understood...this is the difference from Wikipedia and this forum...where the earlier is generalised...you cant clear your doubts...no interaction...Great help sir...I continue to ask whenever the need arises...thank you sir again and thank you MHP...
• August 10th 2010, 10:57 PM
Vlasev
Hehe, the usual procedure is that you click "thanks" whenever you find a post particularly helpful. I agree, sometimes there is no substitute for human interaction. Encyclopedias like Wikipedia tend to be either too simplistic or pretty advanced.