#### HelpNeeder

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and if

A * B = C

can

A * D = C ?

Thank you in advance.

- Thread starter HelpNeeder
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and if

A * B = C

can

A * D = C ?

Thank you in advance.

(if that's what you mean).

All you need is that \(\displaystyle \text{A}\boldsymbol{x}=\boldsymbol{c_j}\) for some column \(\displaystyle \boldsymbol{c_j}\) of C, to have non-unique solutions for \(\displaystyle \boldsymbol{x}\).

E.g.

\(\displaystyle \left(\begin{matrix}2&1&3\\0&1&2\\\end{matrix}\right)\left(\begin{matrix}1\\2\\3\\\end{matrix}\right)=\left(\begin{matrix}13\\8\\\end{matrix}\right)\)

\(\displaystyle \left(\begin{matrix}2&1&3\\0&1&2\\\end{matrix}\right)\left(\begin{matrix}2\\6\\1\\\end{matrix}\right)=\left(\begin{matrix}13\\8\\\end{matrix}\right)\)

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is false.

So I do not understand why they are talking about uniqueness here.

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is true...

I take it you are linking this to your original post:Thank you! The thing is that I do not understand why in this case the solution is unique:

View attachment 26285

Here is the proof:

View attachment 26286

Becayse if I multiply A^{-1}to anotherb, then I can still have the sameu?

Can there exist matrices A, B, C, D:

B ≠ D and AB = C and AD = C

As I showed, such things

However it does not mean that for

In particular, if A is invertible: then for matrix C, there is a unique matrix B: AB = C (namely A

(and so there is a unique solution to A

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Thank you! But how can we be sure that A^{-1}C is unique?

AB = C \(\displaystyle \quad\)(1)Thank you! But how can we be sure that A-1C is unique?

As A is invertible, \(\displaystyle \text{A}^{-1}\) exists.

You can then left multiply both sides of equation (1) by \(\displaystyle \text{A}^{-1}\)

\(\displaystyle \text{A}^{-1}\text{AB} = \text{A}^{-1} \text{C}\)

\(\displaystyle \text{IB}\) = \(\displaystyle \text{A}^{-1}\text{C}\)

\(\displaystyle \text{B=}\text{A}^{-1}\text{C}\)

\(\displaystyle \text{A}^{-1}\text{C}\) is the product of two matrices. You only get one unique answer when you multiply two matrices together.

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Thank you!

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I know that Lex has already told you why AThank you! But how can we be sure that A^{-1}C is unique?