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

Thanks

- Thread starter flash
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- #1

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

Thanks

- #2

tiny-tim

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Hi flash!

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

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

- #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?

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?

CAx = Cb

Ix = Cb

x = Cb

Am I on the right track?

- #6

tiny-tim

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

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.

- #7

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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.

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- #9

HallsofIvy

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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.

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.

(I

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- #10

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I apologize for that. I didn't notice my error about the invertibility.(IthinkBryanP'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 isnotinvertible. Only square matrices are invertible.)

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