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Matrix problem

  • Thread starter flash
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  • #1
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Suppose A is a 3 x 4 matrix and there exists a 4 x 3 matrix C such that AC = I (the 3x3 identity matrix). Let b be an arbitrary vector in R3. Produce a solution of Ax=b.

I'm not quite sure what the question is asking. I think I just need someone to point me in the right direction.

Thanks
 

Answers and Replies

  • #2
tiny-tim
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Hi flash! :smile:

Hint: AC = I … so multiply something by I ! :wink:
 
  • #3
Defennder
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Further hint if the one given above is too vague: Multiply it on the right side of I.
 
  • #4
HallsofIvy
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If AC= I then CA= I.

How would you solve Ax= b if A, x, and b were NUMBERS?
 
  • #5
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Ax = b
CAx = Cb
Ix = Cb
x = Cb

Am I on the right track?
 
  • #6
tiny-tim
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Am I on the right track?
Not only on the right track … you've arrived at Grand Central! :biggrin:

You have proved that A sends Cb to ACb = Ib = b.

So A(Cb) = b.

In other words, x = Cb is a solution to Ax = b. :smile:
 
  • #7
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Ok, thanks. The other part of the question goes:
A is a 4x3 matrix
C is a 3x4 matrix such that CA = I
Suppose, for some given b in R4 that Ax=b has at least one solution. Show that this solution is unique.

Can I just say x = Cb which implies that there is only one solution for x? I'm thinking that I should say something along the lines of: if there exists a C such that CA = I then A must have no free variables.
 
  • #8
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Hint: What does it mean for a matrix to be invertible? If you had an invertible matrix, what does this say about the number of solutions for every b in Ax = b?
 
  • #9
HallsofIvy
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Ok, thanks. The other part of the question goes:
A is a 4x3 matrix
C is a 3x4 matrix such that CA = I
Suppose, for some given b in R4 that Ax=b has at least one solution. Show that this solution is unique.

Can I just say x = Cb which implies that there is only one solution for x? I'm thinking that I should say something along the lines of: if there exists a C such that CA = I then A must have no free variables.
Yes, that's all you need to do. C is a given matrix , b is a given vector: since multiplication of a 3x4 matrix with a 4 dimensional vector (a 4x1 matrix) is "well defined", x= Cb must be a specific, unique vector.

(I think BryanP's response is to your previous question- though then I don't know why he refers to " the number of solutions for every b in Ax = b". Here, A is not invertible. Only square matrices are invertible.)
 
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  • #10
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(I think BryanP's response is to your previous question- though then I don't know why he refers to " the number of solutions for every b in Ax = b". Here, A is not invertible. Only square matrices are invertible.)
I apologize for that. I didn't notice my error about the invertibility.
 

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