Matrix multiplication

HelpNeeder

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If A, B, C and D are arbitrary matrices (with the correct amount of rown and columns to be multiplied)

and if

A * B = C

can

A * D = C ?

Thank you in advance.
 

lex

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There exist matrices A, B, C, D: B ≠ D and AB = C = AD
(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.
\(\displaystyle \left(\begin{matrix}2&1&3\\0&1&2\\\end{matrix}\right)\left(\begin{matrix}1\\2\\3\\\end{matrix}\right)=\left(\begin{matrix}13\\8\\\end{matrix}\right)\)
\(\displaystyle \left(\begin{matrix}2&1&3\\0&1&2\\\end{matrix}\right)\left(\begin{matrix}2\\6\\1\\\end{matrix}\right)=\left(\begin{matrix}13\\8\\\end{matrix}\right)\)
 

HelpNeeder

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Thank you! The thing is that I do not understand why in this case the solution is unique:
1617889046366.png

Here is the proof:
1617889090823.png
Becayse if I multiply A-1 to another b, then I can still have the same u?
 

HelpNeeder

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And the statement
If there is a b in Rn such that the equation Ax = b is consistent, then the solution is unique.
is false.
So I do not understand why they are talking about uniqueness here.
 

HelpNeeder

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And the statement
If the equation Ax = b has at least one solution for each b in Rn, then the solution is unique for each b.
is true...
 

lex

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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 another b, then I can still have the same u?
I take it you are linking this to your original post:
Can there exist matrices A, B, C, D:
B ≠ D and AB = C and AD = C
As I showed, such things do exist.
However it does not mean that for all matrices A, C (of correct sizes), there always exist B ≠ D: AB = C and AD = C

In particular, if A is invertible: then for matrix C, there is a unique matrix B: AB = C (namely A-1C)
(and so there is a unique solution to Ax=b (namely A-1b)).
 

HelpNeeder

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

lex

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

HelpNeeder

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Thank you!
 

Jomo

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Thank you! But how can we be sure that A-1C is unique?
I know that Lex has already told you why A-1C is unique but I have to repeat the answer. Whenever you multiply two matrices of proper sizes you only get one answer!!!!
 
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