A manipulative matrix question

  • Thread starter Raghav Gupta
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In summary: I was saying,If I use the above matrices,##\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]####\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]##I get the following:A = ##\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]##B = ##\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]##A2 = A2BAAB =
  • #1
Raghav Gupta
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Homework Statement



If A and B are 2 matrices such that AB = A and BA =B, then B2 is equal to
B
A
Zero matrix
I

Homework Equations


We can pre or post multiply a matrix on both sides of equation.

The Attempt at a Solution


(AB).(BA) = A.B
AB2A = A.B
Pre multiply both sides by A-1
We get B2A = AB
Implies B2AB = AB as (A = AB)
Post multiplying by B-1 and A we get B2 = I

Is this correct?
 
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  • #2
Raghav Gupta said:

Homework Statement



If A and B are 2 matrices such that AB = A and BA =B, then B2 is equal to
B
A
Zero matrix
I

Homework Equations


We can pre or post multiply a matrix on both sides of equation.

The Attempt at a Solution


(AB).(BA) = A.B
AB2A = A.B
Pre multiply both sides by A-1
We get B2A = AB
Shouldn't that last equation be as follows?

B2A = B

By the way: Are you certain that matrix, A, has an inverse?
Implies B2AB = AB as (A = AB)
Post multiplying by B-1 and A we get B2 = I

Is this correct?
Are you certain that matrix, A, has an inverse?
 
  • #3
I don't see anything in the problem that says that either matrix is invertible, so there's no justification for assuming that A-1 exists.

I found it helpful to start with B2 and work with that.
 
  • #4
Maybe it was a trick question - nothing involving inverse matrices. Both A and B could be identity matrices. This would satisfy the problem statement in the strictest sense.
 
  • #5
Mark44 said:
I found it helpful to start with B2 and work with that.
I started with A2 .
 
  • #6
SammyS said:
I started with A2 .
That seems the long way around if the goal it to find B2, but then I don't know what you did.
 
  • #7
Mark44 said:
That seems the long way around if the goal it to find B2, but then I don't know what you did.
Well, I suppose we'll have to see what OP does before we can all compare notes.
 
  • #8
If you start with ##B^2##,
Then i got A=I.
##B^2=B.B=B.BA=B^2A##
##B^2=B^2A##
A=I satisfies the above equation.
And if i put it on AB=A, then i got B=I.
 
  • #9
AdityaDev said:
If you start with ##B^2##,
Then i got A=I.
##B^2=B.B=B.BA=B^2A##
##B^2=B^2A##
A=I satisfies the above equation.
And if i put it on AB=A, then i got B=I.
While I satisfied that equation, it's not necessary that A = I .
 
  • #10
AdityaDev said:
If you start with ##B^2##,
Then i got A=I.
##B^2=B.B=B.BA=B^2A##
##B^2=B^2A##
A=I satisfies the above equation.
And if i put it on AB=A, then i got B=I.
As Sammy said, B isn't necessarily I. Here's a counterexample, using your assumption that A = I.
$$B = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$
This matrix satisfies B2 = B2A = B2I, yet ##B \neq I##.
 
  • #11
SammyS said:
Shouldn't that last equation be as follows?

B2A = B (Yes, sorry )

By the way: Are you certain that matrix, A, has an inverse? (No)

Are you certain that matrix, A, has an inverse? (No, as Mark44 10th post example proves that matrix B has determinant=0)
So,
I started with B2 ,

## B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B ## Is multiplication of matrices associative which I am doing here?

Starting with A2.
Following the above similar pattern , I get A2= A.
How did you Sammy get B2 value by starting from A2?
 
  • #12
Raghav Gupta said:
So,
I started with B2 ,

## B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B ## Is multiplication of matrices associative which I am doing here?

Starting with A2.
Following the above similar pattern , I get A2= A.
How did you Sammy get B2 value by starting from A2?

Yes, multiplication of matrices is certainly associative.
 
  • #13
Dick said:
Yes, multiplication of matrices is certainly associative.
How?
 
  • #16
Raghav Gupta said:
Oh, I was expecting a easy proof for it. I think we have to go in deep for that topic to understand the proof?

It's not deep, you just have to pay attention to the indices. Ignore the ring stuff. Just use that for real numbers (ab)c=a(bc).
 
  • #17
Dick said:
It's not deep, you just have to pay attention to the indices. Ignore the ring stuff. Just use that for real numbers (ab)c=a(bc).
Thanks, got it.
 
  • #18
Raghav Gupta said:
So,
I started with B2 ,

## B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B ## Is multiplication of matrices associative which I am doing here?

Starting with A2.
Following the above similar pattern , I get A2= A.
How did you Sammy get B2 value by starting from A2?
Of course, I didn't follow that pattern, but, yes, A2= A.

I start with:
A2 = A2

BAAB = BAAB

((BA)A)B = (BA)(AB)

(BA)B = BA

BB = BAdded in Edit:
Here are two such matrices.
##\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]##

##\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]##
 
Last edited:
  • #19
SammyS said:
I start with:
A2 = A2

BAAB = BAAB
How you have written A2 = BAAB ?
 
  • #20
Here are two such matrices.
##\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]##

##\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]##
Raghav Gupta said:
How you have written A2 = BAAB ?
I didn't say that.

I said

BAAB = BAABNow that I look at that, I only needed:

BAB = BAB

(BA)B = B(AB)

BB = BA

B2 = B
 
  • #21
SammyS said:
BAAB = BAABNow that I look at that, I only needed:

BAB = BAB
How you have written A2 = A ?
 
  • #22
Raghav Gupta said:
How you have written A2 = A ?
I didn't use that at all. (I did mention in Post #18 that it is also true.)That post (#20) says, instead of starting with the identity BAAB = BAAB,

I could have started with BAB = BAB .

Either of these equations is obviously true.

You seem to not be following my method of working with these equations.

I manipulate each side of the equation independently.

I could have also done it this way.

BAB = (BA)B
= BB
=B2

BAB = B(AB)
=BA
=B​

Therefore, B2 = B .

Added In Edit:
I left the 2 out of the [ SUP][ /SUP]. DUH !

(Well the little icon has X2, so why no 2 ?) :smile:
 
Last edited:
  • #23
SammyS said:
I was simply stating that similar to B2 = B, it's also true that A2 = A.
I know that already and I have solved it as you can see in my earlier post.
What I am asking is you have told that earlier that you started with A2 and when deriving that value of B2 from A2 you don't know that A2 = A.
So how from A2
BAAB comes?
 
  • #24
SammyS said:
I didn't use that at all. (I did mention in Post #18 that it is also true.)That post (#20) says, instead of starting with the identity BAAB = BAAB,

I could have started with BAB = BAB .

Either of these equations is obviously true.

You seem to not be following my method of working with these equations.

I manipulate each side of the equation independently.

I could have also done it this way.

BAB = (BA)B
= BB
=B2

BAB = B(AB)
=BA
=B​

Therefore, B = B .
I'm pretty sure that you meant Therefore, B2 = B.
 
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  • #25
Raghav Gupta said:
I know that already and I have solved it as you can see in my earlier post.
What I am asking is you have told that earlier that you started with A2 and when deriving that value of B2 from A2 you don't know that A2 = A.
So how from A2
BAAB comes?
I said I started with A2, then I revised that (in the post) to say Instead start with BAAB. (Of course that's BA2B .)

After that, I further made the observation that I might just have well started with BAB .

I don't use A2 = A at all.
 
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  • #26
Okay, got it.
Thanks to all of you.
Learnt quite a lesson from this question.
 
  • #27
Raghav Gupta said:
Okay, got it.
Thanks to all of you.
Learnt quite a lesson from this question.

The easiest solution of all is to start with AB = A and BA =B, and multiply both sides of the first equation on the left by B, to get B(AB) = BA, or (BA)B = B = B*B because BA=B. That seems to me to be about the simplest solution possible.
 
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  • #28
Thanks to you as well Ray.
 

Related to A manipulative matrix question

1. What is a manipulative matrix question?

A manipulative matrix question is a type of survey question that presents a matrix or grid of answer options with corresponding statements or questions. This format allows for more complex and detailed responses, as participants can choose from multiple answer choices for each statement.

2. How is a manipulative matrix question different from other types of survey questions?

Unlike multiple choice or open-ended questions, a manipulative matrix question allows for a combination of both types of responses. It also allows for more precise and specific feedback, as participants can choose from a range of answer options for each statement.

3. What are the benefits of using a manipulative matrix question?

Using a manipulative matrix question can help gather more detailed and accurate data, as it allows participants to give multiple responses to a statement or question. It also helps to organize and categorize responses, making it easier to analyze the data.

4. How can I create an effective manipulative matrix question?

To create an effective manipulative matrix question, it is important to use clear and concise statements or questions that are relevant to the topic. It is also important to provide a balanced range of answer options and to avoid leading or biased statements.

5. In what situations is a manipulative matrix question most useful?

A manipulative matrix question is most useful in situations where you want to gather detailed and specific feedback from participants. It can be particularly useful in market research, customer satisfaction surveys, and employee feedback surveys.

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