Solving Ax=b with Matrices A and C

[flash]

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

 

[tiny-tim]

Hi flash!

Hint: AC = I … so multiply something by I !

 

[Defennder]

Further hint if the one given above is too vague: Multiply it on the right side of I.

 

[HallsofIvy]

If AC= I then CA= I.

How would you solve Ax= b if A, x, and b were NUMBERS?

 

[flash]

Ax = b
CAx = Cb
Ix = Cb
x = Cb

Am I on the right track?

 

[tiny-tim]

Am I on the right track?

Not only on the right track … you've arrived at Grand Central!

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.

 

[flash]

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.

 

[BryanP]

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?

 

[HallsofIvy]

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.)

 

[BryanP]

(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.